Problem: Process the Q queries below.

* You are given two integers A_i and M_i. Determine whether there exists a positive integer K_i not exceeding 2 × 10^{18} such that A_i^{K_i} ≡ K_i (mod M_i), and find one if it exists.

Constraints

* 1 \leq Q \leq 100
* 0 \leq A_i \leq 10^9(1 \leq i \leq Q)
* 1 \leq M_i \leq 10^9(1 \leq i \leq Q)



Examples

Input

4
2 4
3 8
9 6
10 7


Output

4
11
9
2


Input

3
177 168
2028 88772
123456789 987654321


Output

7953
234831584
471523108231963269

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<string>
#include <iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
#define lscan(x) scanf("%I64d",&x);
#define lprint(x) printf("%I64d",x);
ll gcd(ll a, ll b) {
	int c = a % b;
	while (c != 0) {
		a = b;
		b = c;
		c = a % b;
	}
	return b;
}
// 返り値: a と b の最大公約数
// ax + by = gcd(a, b) を満たす (x, y) が格納される
long long extGCD(long long a, long long b, long long &x, long long &y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	long long d = extGCD(b, a%b, y, x);
	y -= a / b * x;
	return d;
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

ll M = 1000000007;

vector<ll> fac(1000001); //n!(mod M)
vector<ll> ifac(1000001); //k!^{M-2} (mod M)

ll mpow(ll x, ll n) {
	ll ans = 1;
	while (n != 0) {
		if (n & 1) ans = ans * x % M;
		x = x * x % M;
		n = n >> 1;
	}
	return ans;
}
ll mpow2(ll x, ll n, ll mod) {
	ll ans = 1;
	while (n != 0) {
		if (n & 1) ans = ans * x % mod;
		x = x * x % mod;
		n = n >> 1;
	}
	return ans;
}
void setcomb() {
	fac[0] = 1;
	ifac[0] = 1;
	for (ll i = 0; i < 1000000; i++) {
		fac[i + 1] = fac[i] * (i + 1) % M; // n!(mod M)
	}
	ifac[1000000] = mpow(fac[1000000], M - 2);
	for (ll i = 1000000; i > 0; i--) {
		ifac[i - 1] = ifac[i] * i%M;
	}
}
ll comb(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	ll tmp = ifac[a - b] * ifac[b] % M;
	return tmp * fac[a] % M;
}
ll perm(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	return fac[a] * ifac[a - b] % M;
}
// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a) {
	long long b = M, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= M;
	if (u < 0) u += M;
	return u;
}
vector<vector<ll>> mul(vector<vector<ll>> a, vector<vector<ll>> b, int n) {
	int i, j, k, t;
	vector<vector<ll>> c(n);
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			t = 0;
			for (k = 0; k < n; k++)
				t = (t + a[i][k] * b[k][j] % M) % M;
			c[i].push_back(t);
		}
	}
	return c;
}

template< typename Monoid >
struct SegmentTree {

	int sz;
	vector< Monoid > seg;

	const Monoid M1;

	SegmentTree(int n, const Monoid &M1) : M1(M1) {
		sz = 1;
		while (sz < n) sz <<= 1;
		seg.assign(2 * sz, M1);
	}

	void set(int k, const Monoid &x) {
		seg[k + sz] = x;
	}

	void build() {
		for (int k = sz - 1; k > 0; k--) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	void update(int k, const Monoid &x) {
		k += sz;
		seg[k] = x;
		while (k >>= 1) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	Monoid query(int a, int b) {
		Monoid L = M1, R = M1;
		for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
			if (a & 1) L = f(L, seg[a++]);
			if (b & 1) R = f(seg[--b], R);
		}
		return f(L, R);
	}

	Monoid value(const int &k) const {
		return seg[k + sz];
	}

	Monoid f(Monoid a, Monoid b) {
		return a + b;
	}
};

ll totient(ll n) {
	ll ans = 1;
	for (ll i = 2; i*i <= n; i++) {
		ll count = 0;
		while (n%i == 0) {
			count++;
			n /= i;
		}
		if (count > 0)
			ans *= mpow(i, count - 1)*(i-1);
	}
	if (n > 1)
		ans *= n - 1;
	return ans;
}

ll solve(ll a, ll m) {
	ll g1 = gcd(a, m);
	ll mp = m; //mp=M'
	while (gcd(g1, mp) > 1)
		mp /= gcd(g1, mp);
	ll to = totient(mp);
	ll g2 = gcd(to, m);
	ll c;
	if (g2 == 1)
		c = 0;
	else
		c = solve(a, g2) % to;
	ll c2 = to;
	c2 *= 100;
	c2 += c;
	ll d = mpow2(a, c2, m);
	ll p = d - c;
	p += m;
	p %= m;
	ll x, y;
	extGCD(to, m, x, y);
	x *= p / g2;
	x %= m;
	ll ans = x * to + c;
	if (ans <= 100)
		ans += ((100-ans)/(to*m)+1)*to*m;
	return ans;
}

int main() {
	ll q,a,m;
	cin >> q;
	for (; q > 0; q--) {
		cin >> a >> m;
		cout << solve(a, m) << endl;
	}
}