Since $$$N$$$ is always even, try visualizing the array as $$$N/2$$$ independent, non-overlapping pairs: $$$(A_1,A_2), (A_3,A_4), \dots$$$

If every pair must sum to a multiple of $$$P$$$, what must the total sum $$$M$$$ be a multiple of? Given that numbers must be positive, what is the absolute minimum possible total sum?

To build the array, try repeating a simple pair like $$$(1,P-1)$$$ for most of the array, and put all the remaining required sum into the very last element.

When to print NO:

Because $$$N$$$ is even, we can split the array into exactly $$$N/2$$$ disjoint pairs. The problem requires every adjacent pair to sum to a multiple of $$$P$$$. Therefore, the sum of the entire array, $$$M$$$, must also be a multiple of $$$P$$$ ($$$M \bmod P = 0$$$).

Additionally, since the array requires strictly positive integers ($$$A_i \ge 1$$$), the smallest valid sum for any pair is $$$P$$$. With $$$N/2$$$ pairs, the absolute minimum total sum is $$$(N/2) \times P$$$.

If $$$M \lt (N/2) \times P$$$, it is impossible.

Therefore, if either of these conditions fails, print NO.

When to print YES:

If both conditions are met, print YES and construct the array greedily:

• Prefix: Fill the first $$$N-2$$$ positions by alternating $$$1$$$ and $$$P-1$$$: $$$1, P-1, 1, P-1, \dots$$$

Every adjacent pair here sums to exactly $$$P$$$.

• Second to last element: Set $$$A_{N-1} = 1$$$.

Since the prefix always ends in $$$P-1$$$, their sum is $$$P-1+1=P$$$, which is valid.

• Final element: Set $$$A_N$$$ to whatever value is needed to make the total array sum exactly $$$M$$$.

Since we already verified that $$$M \ge (N/2)\times P$$$, $$$A_N$$$ will always be a valid positive integer.

#include <bits/stdc++.h>
using namespace std;
 
#define endl '\n'
 
void solve()
{
    int n, m, p;
    cin >> n >> m >> p;
    if ((m % p != 0) || ((m / p) < (n / 2)))
    {
        cout << "NO" << endl;
    }
    else
    {
        cout << "YES" << endl;
        for (int i = 0; i < n - 2; i += 2)
        {
            cout << 1 << " " << p - 1 << " ";
            m -= p;
        }
        cout << 1 << " " << m - 1 << endl;
    }
}
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    int t = 1;
 
    cin >> t;
    while (t--)
    {
        solve();
    }
 
    return 0;
}