#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target ("avx")

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> void chmin(T &t, const T &f) { if (t > f) t = f; }
template <class T> void chmax(T &t, const T &f) { if (t < f) t = f; }


constexpr int LIM = 2005;

bool isnp[100010];
int primesLen;
int primes[320];

int N;
int A[100010];
int Q;
int M[100010], L[100010], R[100010];

int sums[320][100010];

int main() {
  for (int p = 2; p < LIM; ++p) {
    if (!isnp[p]) {
      for (int n = 2 * p; n < LIM; n += p) {
        isnp[n] = true;
      }
    }
  }
  for (int p = 2; p < LIM; ++p) {
    if (!isnp[p]) {
      primes[primesLen++] = p;
    }
  }
  cerr << "|primes| = " << primesLen << endl;
  
  for (; ~scanf("%d", &N); ) {
    for (int i = 0; i < N; ++i) {
      scanf("%d", &A[i]);
    }
    scanf("%d", &Q);
    for (int q = 0; q < Q; ++q) {
      scanf("%d%d%d", &M[q], &L[q], &R[q]);
      --L[q];
    }
    
    for (int h = 0; h < primesLen; ++h) {
      const int p = primes[h];
      for (int i = 0; i < N; ++i) {
        int e = 0;
        for (int a = A[i]; a % p == 0; a /= p) {
          ++e;
        }
        sums[h][i + 1] = sums[h][i] + e;
      }
    }
    
    for (int q = 0; q < Q; ++q) {
      bool ans = true;
      int m = M[q];
      for (int h = 0; h < primesLen; ++h) {
        const int p = primes[h];
        int e = 0;
        for (; m % p == 0; m /= p) {
          ++e;
        }
        ans = ans && (sums[h][R[q]] - sums[h][L[q]] >= e);
      }
      ans = ans && (m == 1);
      puts(ans ? "Yes" : "NO");
    }
    
  }
  
  return 0;
}