#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
int ksm(int a, int b) {
  int s = 1, m = a;
  while (b) {
    if (b & 1) s = 1ll * s * m % mod;
    m = 1ll * m * m % mod, b >>= 1;
  }
  return s;
}
int inv(int x) { return ksm(x, mod - 2); }
int n, q, a[N], d[N << 3], pre[N], ans[N];
vector<pair<int, int> > ed[N];
int c[N];
void add(int x, int v) {
  while (x <= n) c[x] = 1ll * c[x] * v % mod, x += x & -x;
}
int query(int x) {
  int ans = 1;
  while (x) ans = 1ll * ans * c[x] % mod, x -= x & -x;
  return ans;
}
void upd(int x, int v) {
  if (d[v]) add(d[v], 1ll * v * inv(v - 1) % mod);
  add(d[v] = x, 1ll * (v - 1) * inv(v) % mod);
}
int main() {
  scanf("%d", &n), pre[0] = 1;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), pre[i] = 1ll * pre[i - 1] * a[i] % mod, c[i] = 1;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    ed[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    int tmp = a[i];
    for (int j = 2; j * j <= tmp; j++)
      if (tmp % j == 0) {
        upd(i, j);
        while (tmp % j == 0) tmp /= j;
      }
    if (tmp > 1) upd(i, tmp);
    for (pair<int, int> it : ed[i])
      ans[it.second] = 1ll * pre[i] * inv(pre[it.first - 1]) % mod * query(i) %
                       mod * inv(query(it.first - 1)) % mod;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma GCC optimize(3)
using namespace std;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
int ksm(int a, int b) {
  int s = 1, m = a;
  while (b) {
    if (b & 1) s = 1ll * s * m % mod;
    m = 1ll * m * m % mod, b >>= 1;
  }
  return s;
}
int inv(int x) { return ksm(x, mod - 2); }
int n, q, a[N], d[N << 3], pre[N], ans[N];
vector<pair<int, int> > ed[N];
int c[N];
void add(int x, int v) {
  while (x <= n) c[x] = 1ll * c[x] * v % mod, x += x & -x;
}
int query(int x) {
  int ans = 1;
  while (x) ans = 1ll * ans * c[x] % mod, x -= x & -x;
  return ans;
}
void upd(int x, int v) {
  if (d[v]) add(d[v], 1ll * v * inv(v - 1) % mod);
  add(d[v] = x, 1ll * (v - 1) * inv(v) % mod);
}
int main() {
  scanf("%d", &n), pre[0] = 1;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), pre[i] = 1ll * pre[i - 1] * a[i] % mod, c[i] = 1;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    ed[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    int tmp = a[i];
    for (int j = 2; j * j <= tmp; j++)
      if (tmp % j == 0) {
        upd(i, j);
        while (tmp % j == 0) tmp /= j;
      }
    if (tmp > 1) upd(i, tmp);
    for (pair<int, int> it : ed[i])
      ans[it.second] = 1ll * pre[i] * inv(pre[it.first - 1]) % mod * query(i) %
                       mod * inv(query(it.first - 1)) % mod;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

```