積性函式太有趣了!
做這個題之前，我們需要掌握一些基本知識。

1.若f(x)為積性函式，那麼滿足f(xy)=f(x)f(y)
2.若f(x)為積性函式，g(x)=∑d|xf(d)，那麼g(x)也為積性函式
3.學會如何線性處理積性函式。
4.對於積性函式，目標就是學會如何求f(pq)，p為質數

回到這個題，我們先考慮求f函式
f(m)=∑m−1a=0∑m−1b=0[gcd(m,ab)!=m]
我們考慮a b的所有組合，一共m2情況
考慮f函式的反面，即f(x)′=∑m−1a=0∑m−1b=0[gcd(m,ab)==m]
若gcd(m,a)=d,那麼如果滿足gcd

#include <map>
#include <set>
#include <list>
 

#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <memory>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cassert>
#include <climits>
#include <cstring>
 

#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define fuck(x) std::cout << "[" << x << "]";
#define FIN freopen("input.txt", "r", stdin);
#define FOUT freopen("output.txt", "w+", stdout);

typedef long long ll;
typedef long double ldb;
typedef std::pair<ll, int> pli;
typedef std::pair<int, int> pii;
typedef unsigned long long ull;
typedef std::pair<double, double> pdd;

const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 1e9 + 7;

const int MX = 4e4 + 5;

int prime[MX], psz;
bool not_prime[MX];

void prepare() {
    not_prime[1] = true;
    for(int i = 2; i < MX; i++) {
        if(!not_prime[i]) {
            prime[++psz] = i;
        }
        for(int j = 1; j <= psz && (ll)i * prime[j] < MX; j++) {
            int x = i * prime[j];
            not_prime[x] = true;
            if(i % prime[j] == 0) break;
        }
    }
}
ull solve(int n) {
    ull g1 = 1, g2 = 1;
    for(int i = 1; i <= psz && (ll)prime[i]*prime[i] <= n; i++) {
        if(n % prime[i] == 0) {
            int p = prime[i], q = 0;
            while(n % prime[i] == 0) {
                q++; n /= prime[i];
            }
            ull f1 = 1, f2 = q + 1;
            for(int j = 1; j <= q; j++) {
                f1 = f1 * p * p + 1;
                f2 = f2 * p;
            }
            g1 *= f1; g2 *= f2;
        }
    }
    if(n != 1) {
        int p = n;
        g1 *= (ull)p * p + 1;
        g2 *= 2 * p;
    }
    return g1 - g2;
}
int main() {
    // FIN;
    prepare();
    int T; scanf("%d", &T);
    while(T--) {
        int n; scanf("%d", &n);
        printf("%llu\n", solve(n));
    }
    return 0;
}