// yukicoder: No.109 N! mod M
// 2019.4.21 bal4u

#include <stdio.h>

//// 高速入力
#if 1
#define gc() getchar_unlocked()
#define pc(c) putchar_unlocked(c)
#else
#define gc() getchar()
#define pc(c) putchar(c)
#endif
int in()    // 非負整数の入力
{
	int n = 0, c = gc();
	do n = 10 * n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

void out(int n)  // 非負整数の表示（出力）
{
	int i;
	char b[20];

	if (!n) pc('0');
	else {
		i = 0; while (n) b[i++] = n % 10 + '0', n /= 10;
		while (i--) pc(b[i]);
	}
	pc('\n');
}


//// 素数テスト
long long calc_pow(int a, int k, int n)
{
	long long p;
	unsigned bit;

	p = 1;
	for (bit = 0x80000000U; bit > 0; bit >>= 1) {
		if (p > 1) p = (p*p) % n;
		if (k & bit) p = (p*a) % n;
	}
	return p % n;
}

int suspect(int b, int n)
{
	int i, t, u, n1;
	long long x0, x1;

	u = n1 = n - 1;
	t = 0;
	while ((u & 1) == 0) {
		t++;
		u >>= 1;
	}
	x0 = calc_pow(b, u, n);
	for (i = 1; i <= t; i++) {
		x1 = (x0*x0) % n;
		if (x1 == 1 && x0 != 1 && x0 != n1) return 0;
		x0 = x1;
	}
	return x1 == 1;
}

int prime_test(int n)
{
	if (n == 1) return 0;          /* 1 */
	if (n == 2 || n == 3) return 1;          /* 2, 3 */
	if ((n & 1) == 0) return 0;  /* other even integers */
	if (!suspect(2, n)) return 0;
	if (n <= 1000000) {
		if (!suspect(3, n)) return 0;
	}
	else {
		if (!suspect(7, n)) return 0;
		if (!suspect(61, n)) return 0;
	}
	return 1;
}


//// 逆数の計算
int extended_gcd(int a, int b, int *x, int *y)
{
	int d;
	if (b == 0) { *x = 1; *y = 0; return a; }
	d = extended_gcd(b, a % b, y, x);
	*y -= a / b * (*x);
	return d;
}

int inverse(int a, int m)
{
    int x, y;
    if (extended_gcd(a, m, &x, &y) == 1)
		return ((long long)x + m) % m;
    return 0;
}


//// 本問題関連
int main()
{
	int i, T, N, M;
	long long a, ans;

	T = in();
	while (T--) {
		N = in(), M = in();
		if (N >= M || M == 1) ans = 0;
		else if (N == 0) ans = 1;
		else if (N <= 100000) {
			ans = 1;
			for (i = 2; i <= N; i++) ans = ans * i % M;
		} else if (!prime_test(M)) ans = 0;
		else {
			a = 1;
			for (i = N+1; i < M; i++) a = a * i % M;
			ans = (long long)(M-1) * inverse((int)a, M) % M;
		}
		printf("%d\n", (int)ans);
	}
	return 0;
}