1 sounds like the minimum value we can get. Why? Is it always true?

When n ≤ 2, we can do nothing. When n ≥ 3, since , n - 1 isn't a divisor of n, so we can choose it and get 1 as the result.

n = int(input())
	
if n == 2:
    print(2)
else:
	print(1)

Think what happens to the bitwise OR of the array elements when you perform the allowed operation. Can the overall OR value ever decrease?

Let m = a[1] | a[2] | ... | a[n].

• This value m is the lower bound of the array sum, since: m ≤ a[1] + a[2] + ... + a[n]
• This value is also achievable using valid operations:
  • Repeatedly pick two elements a[i], a[j]
  • Replace one of them with 0, and the other with a[i] | a[j] (which preserves the total OR)
  • Keep doing this until all but one element are 0, and one element holds the total OR m.
So, no matter how you do the operations, you can always reduce the array to [0, 0, ..., m], giving you a total sum of m.

Time Complexity: O(N)

t = int(input())
for _ in range(t):
   n = int(input())
   nums = [int(i) for i in input().split()]
   
   answer = 0
   for num in nums:
      answer |= num
            
   print(answer)

Try to compute the result of ai | z for each element.
You’ll notice z only gets worse (decreases or stays the same) with every operation.

You're given an array a and an integer z. You can perform the following operation any number of times:

• Pick an index i (1 ≤ i ≤ n)
• Set a[i] = a[i] | z
• Set z = a[i] & z

The key insight lies in understanding how the value of z changes after each operation.

Observations:

1. z is Monotonically Decreasing (Bitwise)

• The operation z = a[i] & z only unsets bits in z.
• Once a bit in z becomes 0, it can never become 1 again.
• This means z is most powerful (has the most 1s) at the beginning, and gets strictly worse the more you use it.

1. Best Use of z is at the Start

• Since z weakens over time, using it even once changes its value forever.
• So the most strategic approach is to try all possible a[i] | z without modifying z, and pick the maximum.

1. Why Not Simulate All Operations?

• With the above point, it becomes clear that using z even once may prevent reaching optimal results later.
• Trying all operation sequences would be overkill and unnecessary.

Approach:

1. For each ai in the array:

• Compute ai | z
• Track the maximum value among these.

1. Output the maximum value for each test case.

Example Walkthrough:

Input:

2 3  
3 4

• 3 | 3 = 3
• 4 | 3 = 7

Max is 7 — That’s our answer.

Complexity Analysis:

• For each test case, we do a single pass through the array: O(n)
• Total operations across all test cases: O(10⁴) (within limits)

t = int(input())
for _ in range(t):
    n, z =  map(int, input().split())
    a = list(map(int, input().split()))
    max_value = 0
    for ai in a:
        max_value = max(max_value, ai | z)
    print(max_value)

First solve a simplified version: suppose that in type $$$1$$$ queries, only a single character of the string s is inverted, i.e., $$$l=r$$$ in all type $$$1$$$ queries.

When we invert $$$s_i$$$, the XOR of all numbers from group $$$0$$$ and group $$$1$$$ change in the same way, regardless of whether this inversion was from $$$0$$$ to $$$1$$$ or from $$$1$$$ to $$$0$$$.

We will store $$$2$$$ variables: $$$X_0,X_1$$$, which represent the XOR of all numbers from group $$$0$$$ and group $$$1$$$, respectively. When answering a query of type $$$2$$$, we will simply output either $$$X_0$$$ or $$$X_1$$$. Now we need to understand how to update $$$X_0$$$ and $$$X_1$$$ after receiving a query of type $$$1$$$.

Let's first solve a simplified version: suppose that in type $$$1$$$ queries, only a single character of the string s is inverted, i.e., $$$l=r$$$ in all type $$$1$$$ queries.

Let's see how $$$X_0$$$ and $$$X_1$$$ change after this query. If $$$s_i$$$ was $$$0$$$ and became $$$1$$$, then the number $$$a_i$$$ will be removed from group $$$0$$$. Since $$$XOR$$$ is its own inverse operation ($$$w⊕w=0$$$), we can do this with $$$X_0=X_0⊕a_i$$$. And in $$$X_1$$$, we need to add the number $$$a_i$$$, since now $$$s_i=1$$$. And we can do this with $$$X_1=X_1⊕a_i$$$.

The same thing happens if $$$s_i$$$ was $$$1$$$.

This is the key observation: when we invert $$$s_i$$$, $$$X_0$$$ and $$$X_1$$$ change in the same way, regardless of whether this inversion was from $$$0$$$ to $$$1$$$ or from $$$1$$$ to $$$0$$$.

Therefore, to update $$$X_0$$$ and $$$X_1$$$ after a query of type $$$1$$$ with parameters $$$l,r,$$$ we need to do this: $$$X_0=X_0⊕(a_l⊕…⊕a_r)$$$, and the same for $$$X_1$$$.

To quickly find the XOR value on a subsegment of the array a, we can use a prefix XOR array. If $$$p_i=a_1⊕…a_i$$$, then: $$$a_l⊕…⊕a_r=p_r⊕p_{l−1}$$$.

import sys
t = int(sys.stdin.readline().strip())
for _ in range(t):
    n = int(sys.stdin.readline().strip())
    a = list(map(int, sys.stdin.readline().strip().split()))
    s = sys.stdin.readline().strip()
    q = int(sys.stdin.readline().strip())

    pref = [0] * (n + 1)
    for i in range(1, n + 1):
        pref[i] = pref[i - 1] ^ a[i-1]

    x_0 = 0
    x_1 = 0
    for i in range(n):
        if s[i] == "0":
            x_0 ^= a[i]
        else:
            x_1 ^= a[i]

    ans = []
    for i in range(q):
        line = list(map(int, sys.stdin.readline().strip().split()))
        if line[0] == 2:
            if line[1]:
                ans.append(int(str(x_1)))
            else:
                ans.append(int(str(x_0)))
        else:
            l, r = line[1], line[2]
            seg = pref[r] ^ pref[l - 1]   
            x_0 ^= seg
            x_1 ^= seg
    print(*ans)

The formula given in this task looks difficult to calculate, so we can rewrite it:

$$$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (x_i \, \& \, x_j) \cdot (x_j \, | \, x_k) = \sum_{j=1}^n \sum_{i=1}^n (x_i \, \& \, x_j) \sum_{k=1}^n (x_j \, | \, x_k) = \sum_{j=1}^n \left[ \sum_{i=1}^n (x_i \, \& \, x_j) \right] \cdot \left[ \sum_{k=1}^n (x_j \, | \, x_k) \right]$$$

We fix the element $$$x_j$$$. Now the task is to calculate two sums $$$\sum_i (x_i \, \& \, x_j)$$$ and $$$\sum_k (x_j \, | \, x_k)$$$, and multiply them by each other.

Let's define function $$$f(x, c)$$$ as the value of $$$c$$$-th bit in $$$x$$$. For example $$$f(13, 1) = 0$$$, because $$$13 = 11\underline{0}1_2$$$, and $$$f(12, 2) = 1$$$, because $$$12 = 1\underline{1}00_2$$$. Additionally, define $$$M$$$ as the smallest integer such that $$$\forall_i \, x_i \lt 2^M$$$. Note that in this task $$$M \leq 60$$$.

We can rewrite our sums using function $$$f$$$:

$$$\sum_i (x_i \, \& \, x_j) = \sum_{c = 0}^{M} 2^c \sum_i f(x_i, c) \cdot f(x_j, c) = \sum_{c = 0}^{M} 2^c f(x_j, c) \sum_i f(x_i, c)$$$

$$$\sum_k (x_j \, | \, x_k) = \sum_{c = 0}^{M} 2^c \sum_k 1 - (1 - f(x_j, c)) \cdot (1 - f(x_k, c)) = \sum_{c = 0}^{M} 2^c \left[ n - (1 - f(x_j, c)) \sum_k (1 - f(x_k, c)) \right]$$$

In other words, we just split elements $$$x_i$$$, $$$x_j$$$, $$$x_k$$$ into the powers of two.

If we memorize the values of $$$\sum_i f(x_i, c)$$$, for each $$$c \in {0, 1, \ldots, M }$$$, then we can calculate the desired sums in $$$\mathcal{O}(M)$$$ for fixed $$$x_j$$$ using the above equations.

So the final solution is to iterate over all elements in the array and fix them as $$$x_j$$$, and sum all of the results obtained. Complexity is $$$\mathcal{O}(nM) = \mathcal{O}(n \log \max_i(x_i))$$$

from sys import stdin


def input(): return stdin.readline().strip()

T = int(input())

for _ in range(T):
    N = int(input())
    x = list(map(int, input().split()))

    MOD = 1000_000_007
    bit_counts = [0]*60
    for xi in x:
        for i in range(60):
            if xi&(1<<i):
                bit_counts[i] += 1
    ans = 0
    for xi in x:
        tot_or = tot_and = 0
        for i in range(60):
            if xi&(1<<i):
                tot_or += (1<<i)%MOD*N
                tot_or %= MOD
                tot_and += (1<<i)%MOD*bit_counts[i]
                tot_and %= MOD
            else:
                tot_or += (1<<i)%MOD*bit_counts[i]
                tot_or %= MOD
        ans += tot_or*tot_and
        ans %= MOD

    print(ans)