#include <bits/stdc++.h>
using namespace std;
long long n, m, a[1000001], t[4000001], now[1000001];
long long divp[1000001];
vector<pair<int, int> > v[1000001];
void upd(int v, int l, int r, int pos, long long val) {
  if (l == r && pos == l) {
    t[v] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    upd(2 * v, l, mid, pos, val);
  else
    upd(2 * v + 1, mid + 1, r, pos, val);
  t[v] = (t[2 * v + 1] * t[2 * v]) % 1000000007;
  return;
}
int get(int v, int l, int r, int l1, int r1) {
  if (l > r1 || r < l1) return 1;
  if (l == l1 && r == r1) {
    return t[v];
  }
  int mid = (l + r) / 2;
  long long A = get(2 * v, l, mid, l1, min(mid, r1));
  long long B = get(2 * v + 1, mid + 1, r, max(mid + 1, l1), r1);
  return A * B % 1000000007;
}
long long l[1000001], ans[500001];
int main() {
  cin >> n;
  for (int k = 2; k <= 1000000; ++k) {
    t[k] = 1;
    if (divp[k] == 0) {
      divp[k] = k;
      for (int i = k + k; i <= 1000000; i += k) divp[i] = k;
    }
  }
  for (int k = 1; k <= n; ++k) cin >> a[k];
  int q;
  cin >> q;
  for (int k = 1; k <= q; ++k) {
    int l, r;
    cin >> l >> r;
    v[r].push_back({l, k});
  }
  for (int k = 1; k <= n; ++k) {
    long long val = a[k], x = a[k];
    while (val > 1) {
      long long d = divp[val];
      if (l[d] != k) {
        x = x / d * (d - 1);
        long long u = now[l[d]] / (d - 1) * d;
        now[l[d]] = u;
        if (l[d] != 0) upd(1, 1, n, l[d], u);
        l[d] = k;
      }
      val /= d;
    }
    now[k] = x;
    upd(1, 1, n, k, x);
    for (int i = 0; i < v[k].size(); ++i) {
      int l = v[k][i].first, ind = v[k][i].second;
      ans[ind] = get(1, 1, n, l, k);
    }
  }
  for (int k = 1; k < q + 1; ++k) cout << ans[k] << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
long long n, m, a[1000001], t[4000001], now[1000001];
long long divp[1000001];
vector<pair<int, int> > v[1000001];
void upd(int v, int l, int r, int pos, long long val) {
  if (l == r && pos == l) {
    t[v] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    upd(2 * v, l, mid, pos, val);
  else
    upd(2 * v + 1, mid + 1, r, pos, val);
  t[v] = (t[2 * v + 1] * t[2 * v]) % 1000000007;
  return;
}
int get(int v, int l, int r, int l1, int r1) {
  if (l > r1 || r < l1) return 1;
  if (l == l1 && r == r1) {
    return t[v];
  }
  int mid = (l + r) / 2;
  long long A = get(2 * v, l, mid, l1, min(mid, r1));
  long long B = get(2 * v + 1, mid + 1, r, max(mid + 1, l1), r1);
  return A * B % 1000000007;
}
long long l[1000001], ans[500001];
int main() {
  cin >> n;
  for (int k = 2; k <= 1000000; ++k) {
    t[k] = 1;
    if (divp[k] == 0) {
      divp[k] = k;
      for (int i = k + k; i <= 1000000; i += k) divp[i] = k;
    }
  }
  for (int k = 1; k <= n; ++k) cin >> a[k];
  int q;
  cin >> q;
  for (int k = 1; k <= q; ++k) {
    int l, r;
    cin >> l >> r;
    v[r].push_back({l, k});
  }
  for (int k = 1; k <= n; ++k) {
    long long val = a[k], x = a[k];
    while (val > 1) {
      long long d = divp[val];
      if (l[d] != k) {
        x = x / d * (d - 1);
        long long u = now[l[d]] / (d - 1) * d;
        now[l[d]] = u;
        if (l[d] != 0) upd(1, 1, n, l[d], u);
        l[d] = k;
      }
      val /= d;
    }
    now[k] = x;
    upd(1, 1, n, k, x);
    for (int i = 0; i < v[k].size(); ++i) {
      int l = v[k][i].first, ind = v[k][i].second;
      ans[ind] = get(1, 1, n, l, k);
    }
  }
  for (int k = 1; k < q + 1; ++k) cout << ans[k] << endl;
  return 0;
}

```