#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7, M = 1e6 + 10;
int last[M], inv[M];
int fen[N], ans[N], a[N];
vector<int> primes[M];
vector<pair<int, int> > que[N];
long long get(int p) {
  long long res = 1;
  for (; p > 0; p -= p & -p) res = (1LL * res * fen[p]) % mod;
  return res;
}
void upd(int p, int vl) {
  for (p++; p < N; p += p & -p) fen[p] = (1LL * fen[p] * vl) % mod;
  return;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = power(a, b / 2);
  res *= res;
  res %= mod;
  if (b & 1) res *= a, res %= mod;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < M; i++) inv[i] = power(i, mod - 2);
  for (int i = 2; i < M; i++)
    if ((int)primes[i].size() == 0)
      for (int j = i; j < M; j += i) primes[j].push_back(i);
  fill(fen, fen + N, 1);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    que[r].push_back({l, i});
  }
  memset(last, -1, sizeof last);
  for (int r = 0; r < n; r++) {
    upd(r, a[r]);
    for (auto p : primes[a[r]]) {
      if (last[p] != -1) upd(last[p], (1LL * p * inv[p - 1]) % mod);
      last[p] = r;
      upd(r, (1LL * (p - 1) * inv[p]) % mod);
    }
    for (auto X : que[r])
      ans[X.second] = (1LL * get(r + 1) * power(get(X.first), mod - 2)) % mod;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}

### Prompt

Please create a solution in cpp to the following problem:

Today on a math lesson the teacher told Vovochka that the Euler function of a positive integer φ(n) is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n. The number 1 is coprime to all the positive integers and φ(1) = 1.

Now the teacher gave Vovochka an array of n positive integers a1, a2, ..., an and a task to process q queries li ri — to calculate and print <image> modulo 109 + 7. As it is too hard for a second grade school student, you've decided to help Vovochka.

Input

The first line of the input contains number n (1 ≤ n ≤ 200 000) — the length of the array given to Vovochka. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106).

The third line contains integer q (1 ≤ q ≤ 200 000) — the number of queries. Next q lines contain the queries, one per line. Each query is defined by the boundaries of the segment li and ri (1 ≤ li ≤ ri ≤ n).

Output

Print q numbers — the value of the Euler function for each query, calculated modulo 109 + 7.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10
7
1 1
3 8
5 6
4 8
8 10
7 9
7 10


Output

1
4608
8
1536
192
144
1152


Input

7
24 63 13 52 6 10 1
6
3 5
4 7
1 7
2 4
3 6
2 6


Output

1248
768
12939264
11232
9984
539136

Note

In the second sample the values are calculated like that:

  * φ(13·52·6) = φ(4056) = 1248
  * φ(52·6·10·1) = φ(3120) = 768
  * φ(24·63·13·52·6·10·1) = φ(61326720) = 12939264
  * φ(63·13·52) = φ(42588) = 11232
  * φ(13·52·6·10) = φ(40560) = 9984
  * φ(63·13·52·6·10) = φ(2555280) = 539136

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7, M = 1e6 + 10;
int last[M], inv[M];
int fen[N], ans[N], a[N];
vector<int> primes[M];
vector<pair<int, int> > que[N];
long long get(int p) {
  long long res = 1;
  for (; p > 0; p -= p & -p) res = (1LL * res * fen[p]) % mod;
  return res;
}
void upd(int p, int vl) {
  for (p++; p < N; p += p & -p) fen[p] = (1LL * fen[p] * vl) % mod;
  return;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = power(a, b / 2);
  res *= res;
  res %= mod;
  if (b & 1) res *= a, res %= mod;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < M; i++) inv[i] = power(i, mod - 2);
  for (int i = 2; i < M; i++)
    if ((int)primes[i].size() == 0)
      for (int j = i; j < M; j += i) primes[j].push_back(i);
  fill(fen, fen + N, 1);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    que[r].push_back({l, i});
  }
  memset(last, -1, sizeof last);
  for (int r = 0; r < n; r++) {
    upd(r, a[r]);
    for (auto p : primes[a[r]]) {
      if (last[p] != -1) upd(last[p], (1LL * p * inv[p - 1]) % mod);
      last[p] = r;
      upd(r, (1LL * (p - 1) * inv[p]) % mod);
    }
    for (auto X : que[r])
      ans[X.second] = (1LL * get(r + 1) * power(get(X.first), mod - 2)) % mod;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}

```