Problem with SPOJ GCDEX

I was solving this problem. The formula for this problem can be founded here. I've stress tested my code, and found out that this is the smallest test case where my code failed (answer for n <= 225537 are all correct).

Input

Output

Expect

It is obvious that my output is wrong, since $$$answer(n) \gt answer(n-1)$$$. I have no idea why my code doesn't work.

Code

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 5;

bool isprime[N];
vector<int> prime;
int mobius[N];
int phi[N];
int phi_pref[N];

void init() {
	fill(isprime, isprime + N, true);
	isprime[0] = isprime[1] = false;
	mobius[1] = 1;
	phi[1] = 1;
	for(int i = 2; i < N; ++i) {
		if(isprime[i]) {
			prime.push_back(i);
			mobius[i] = -1;
			phi[i] = i - 1;
		}
		for(auto& p : prime) {
			if(p * i >= N) {
				break;
			}
			isprime[p * i] = false;
			mobius[p * i] = mobius[p] * mobius[i];
			phi[p * i] = phi[p] * phi[i];
			if(i % p == 0) {
				mobius[p * i] = 0;
				phi[p * i] = phi[i] * p;
				break;
			}
		}
	}
	for(int i = 1; i < N; ++i) {
		phi_pref[i] = phi_pref[i - 1] + phi[i];
	}
}

int main(int argc, char* argv[]) {
	ios::sync_with_stdio(false);
	cin.tie(0);
	init();
	int n;
	while(cin >> n && n) {
		long long ans = 0;
		for(int i = 1, j; i <= n; i = j) {
			int p = n / i;
			j = n / p + 1;
			ans += 1LL * p * (p - 1) / 2 * (phi_pref[j - 1] - phi_pref[i - 1]);
		}
		cout << ans << "\n";
	}
	return 0;
}