#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>

using namespace std;

#define ll long long 
#define MAX_P 1000000

ll M, A, B, fact[30];
int p2[20], fprim[80000];
bool prim[MAX_P];

void prec() {
	//precalculez puterile lui 2 pana la 2 ^ 20, si numerele prime mai mici ca sqrt(B), adica mai mici ca 1 000 000
	for (int i = 0; i < 20; i++) p2[i] = 1 << i;

	fill(prim + 2, prim + MAX_P, 1);

	for (int i = 2; i < MAX_P; i++)
		if (prim[i]) {
			for (int j = 2 * i; j < MAX_P; j += i)
				prim[j] = false;
			fprim[++fprim[0]] = i;
		}
}

void solve() {
    ll t = 0, d = 0;
    while (B > 1) {
		d++;
        if (B % fprim[d] == 0) {
            fact[++t] = fprim[d];
            while (B % fprim[d] == 0)
                B /= fprim[d];
        }

        if (fprim[d] > sqrt(B) && B > 1) {
            fact[++t] = B;
            B = 1;
        }
    }

    ll sol = A;
    for (int i = 1; i < (1 << t); i++) {
        ll nr = 0, prod = 1;
        for (int j = 0; j < t; j++)
            if ((1 << j) & i) {
                prod = 1LL * prod * fact[j + 1];
                nr++;
            }

        if (nr % 2) nr = -1;
        else nr = 1;

        sol = sol + 1LL * nr * A / prod;
    }

    printf("%lld\n", sol);
}

int main() {

    freopen("pinex.in", "r", stdin);
    freopen("pinex.out", "w", stdout); 

	prec();

    scanf("%lld", &M);
   	for (int i = 1; i <= M; i++) {
        scanf("%lld %lld", &A, &B);
   	    solve();
    }

    return 0;
}