#include <bits/stdc++.h>
using namespace std;
struct ask {
  int l, r, where;
} q[210000];
const int mo = 1e9 + 7, N = 1000000;
int A[210000], x[210000], pd[1100000], n, m, B[210000], where[1100000],
    ans[210000];
vector<int> p[1100000];
int quick(int k1, int k2) {
  int k3 = 1;
  while (k2) {
    if (k2 & 1) k3 = 1ll * k3 * k1 % mo;
    k2 >>= 1;
    k1 = 1ll * k1 * k1 % mo;
  }
  return k3;
}
int compare(ask k1, ask k2) { return k1.r < k2.r; }
void add(int k1, int k2) {
  if (k1 == 0) return;
  for (; k1 <= n; k1 += k1 & (-k1)) B[k1] = 1ll * B[k1] * k2 % mo;
}
int find(int k1) {
  int ans = 1;
  for (; k1; k1 -= k1 & (-k1)) ans = 1ll * ans * B[k1] % mo;
  return ans;
}
void addin(int k1, int k2) {
  int w = 1ll * (k1 - 1) * quick(k1, mo - 2) % mo;
  add(where[k1], quick(w, mo - 2));
  add(k2, w);
  where[k1] = k2;
}
void insert(int k1) {
  int w = x[k1];
  for (int i = 0; i < p[w].size(); i++) {
    int k2 = p[w][i];
    addin(k2, k1);
  }
}
int getans(int L, int R) {
  return 1ll * A[R] * quick(A[L - 1], mo - 2) % mo * find(R) % mo *
         quick(find(L - 1), mo - 2) % mo;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  scanf("%d", &m);
  ;
  A[0] = 1;
  for (int i = 1; i <= n; i++) A[i] = 1ll * A[i - 1] * x[i] % mo;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].where = i;
  }
  sort(q + 1, q + m + 1, compare);
  for (int i = 1; i <= n; i++) B[i] = 1;
  for (int i = 2; i <= N; i++)
    if (pd[i] == 0)
      for (int j = i; j <= N; j += i) {
        pd[j] = 1;
        p[j].push_back(i);
      }
  int now = 1;
  for (int i = 1; i <= n; i++) {
    insert(i);
    while (now <= m && q[now].r == i)
      ans[q[now].where] = getans(q[now].l, q[now].r), now++;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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;
struct ask {
  int l, r, where;
} q[210000];
const int mo = 1e9 + 7, N = 1000000;
int A[210000], x[210000], pd[1100000], n, m, B[210000], where[1100000],
    ans[210000];
vector<int> p[1100000];
int quick(int k1, int k2) {
  int k3 = 1;
  while (k2) {
    if (k2 & 1) k3 = 1ll * k3 * k1 % mo;
    k2 >>= 1;
    k1 = 1ll * k1 * k1 % mo;
  }
  return k3;
}
int compare(ask k1, ask k2) { return k1.r < k2.r; }
void add(int k1, int k2) {
  if (k1 == 0) return;
  for (; k1 <= n; k1 += k1 & (-k1)) B[k1] = 1ll * B[k1] * k2 % mo;
}
int find(int k1) {
  int ans = 1;
  for (; k1; k1 -= k1 & (-k1)) ans = 1ll * ans * B[k1] % mo;
  return ans;
}
void addin(int k1, int k2) {
  int w = 1ll * (k1 - 1) * quick(k1, mo - 2) % mo;
  add(where[k1], quick(w, mo - 2));
  add(k2, w);
  where[k1] = k2;
}
void insert(int k1) {
  int w = x[k1];
  for (int i = 0; i < p[w].size(); i++) {
    int k2 = p[w][i];
    addin(k2, k1);
  }
}
int getans(int L, int R) {
  return 1ll * A[R] * quick(A[L - 1], mo - 2) % mo * find(R) % mo *
         quick(find(L - 1), mo - 2) % mo;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  scanf("%d", &m);
  ;
  A[0] = 1;
  for (int i = 1; i <= n; i++) A[i] = 1ll * A[i - 1] * x[i] % mo;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].where = i;
  }
  sort(q + 1, q + m + 1, compare);
  for (int i = 1; i <= n; i++) B[i] = 1;
  for (int i = 2; i <= N; i++)
    if (pd[i] == 0)
      for (int j = i; j <= N; j += i) {
        pd[j] = 1;
        p[j].push_back(i);
      }
  int now = 1;
  for (int i = 1; i <= n; i++) {
    insert(i);
    while (now <= m && q[now].r == i)
      ans[q[now].where] = getans(q[now].l, q[now].r), now++;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```