// Arup Guha
// 10/12/2023
// Solution to CIS 3362 Homework #5 Question #9

using namespace std;

#include <iostream>
#include <vector>

int phi(int n);
long long sum(int n);

int main() {

    int nC;
    cin >> nC;

    // Process cases.
    for (int loop=0; loop<nC; loop++) {
        int p;
        cin >> p;
        cout << sum(p-1) << endl;
    }

    return 0;
}

// n is guaranteed to be p-1 for some prime p.
long long sum(int n) {

    long long res = 0;

    // Look for all divisors.
    for (int i=1; i*i<=n; i++) {

        // We found 2 divisors, potentially: i and n/i.
        if (n%i == 0) {

            // This is the contribution for i.
            res += ((long long)i)*phi(i);

            // This is the contribution for n/i.
            if (n/i > i)
                res += ((long long)(n/i))*phi(n/i);
        }
    }

    // Ta da!
    return res;
}

// Returns phi(n);
int phi(int n) {

    // We will multiply res by terms of the form (p-1)/p for each unique prime divisor, p of n.
    int res = n;

    // Go to the square root.
    for (int i=2; i*i<=n; i++) {

        // See how manyy times i divides into n and divide it out.
        int exp = 0;
        while (n%i == 0) {
            exp++;
            n /= i;
        }

        // Multiply into res, since res is divisible by i, no inaccuracy occurs.
        if (exp) res = res/i*(i-1);
    }

    // Just in case we have a prime leftover.
    if (n > 1) res = res/n*(n-1);
    return res;
}