#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)2e18;
const int MAXN = (int)1e4 + 5;
const int P = (int)3e7 + (int)3e6;
const int M = (int)1e5;
pair<pair<long long, long long>, int> req[MAXN];
vector<int> primes;
bool isP[P + 5];
bool ans[MAXN];
long long dp[M + 5];
int q;
void pre() {
  for (int i = 2; i <= P; ++i) {
    if (!isP[i]) {
      primes.push_back(i);
      if (i * 1ll * i <= P) {
        for (int j = i * i; j <= P; j += i) {
          isP[j] = 1;
        }
      }
    }
  }
}
long long binPow(long long a, long long b, long long m) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) {
      ret = (ret * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return ret;
}
vector<long long> getPrimes(long long x) {
  vector<long long> ret;
  for (int p : primes) {
    if (x % p == 0) {
      ret.push_back(p);
      while (x % p == 0) {
        x /= p;
      }
    }
  }
  if (x > 1) {
    ret.push_back(x);
  }
  return ret;
}
void calcDp(vector<long long> V) {
  for (long long i = 0; i < V[0]; ++i) {
    dp[i] = INF;
  }
  dp[0] = 0;
  set<pair<long long, int>> S;
  S.insert(make_pair(0, 0));
  while (!S.empty()) {
    int v = S.begin()->second;
    S.erase(S.begin());
    for (long long cur : V) {
      long long to = (v + cur) % V[0];
      if (dp[v] + cur < dp[to]) {
        S.erase(make_pair(dp[to], to));
        dp[to] = dp[v] + cur;
        S.insert(make_pair(dp[to], to));
      }
    }
  }
}
void solve() {
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%lld %lld", &req[i].first.second, &req[i].first.first);
    req[i].second = i;
  }
  sort(req + 1, req + q + 1);
  vector<long long> V;
  long long cur = 0;
  for (int i = 1; i <= q; ++i) {
    long long n = req[i].first.second;
    long long k = req[i].first.first;
    int id = req[i].second;
    if (cur != k) {
      cur = k;
      V = getPrimes(cur);
      if (V.size() > 2) {
        calcDp(V);
      }
    }
    if (cur == 1) {
      ans[id] = 0;
      continue;
    }
    if (V.size() == 1) {
      ans[id] = (n % k == 0);
    } else if (V.size() == 2) {
      long long p = V[0], q = V[1];
      long long x = ((n % q) * binPow(p, q - 2, q)) % q;
      if (n - x * p >= 0) {
        ans[id] = 1;
      }
    } else {
      if (dp[n % V[0]] <= n) {
        ans[id] = 1;
      }
    }
  }
  for (int i = 1; i <= q; ++i) {
    printf(ans[i] ? "YES\n" : "NO\n");
  }
}
int main() {
  int tt = 1;
  pre();
  while (tt--) {
    solve();
  }
  return 0;
}
