#include <bits/stdc++.h>
using namespace std;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int p[1000005], inv[1000005];
long long fastPow(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ans;
}
void init() {
  for (int i = 2; i < 1000005; i++)
    if (!p[i]) {
      for (int j = i; j < 1000005; j += i) {
        p[j] = i;
      }
    }
  inv[0] = 1;
  for (int(i) = 1; (i) <= (int)(1000005 - 1); (i)++)
    inv[i] = fastPow(i, 1000000007 - 2);
}
int a[1000005];
int b[1000005], nxt[1000005];
void update(int x, long long val) {
  for (int i = x; i <= n; i += i & -i) {
    b[i] = (long long)b[i] * val % 1000000007;
  }
}
int query(int x) {
  int ret = 1;
  for (; x > 0; ret = (long long)ret * b[x] % 1000000007, x -= x & -x)
    ;
  return ret;
}
vector<pair<int, int> > ap[1000005];
int ans[1000005];
void fmain() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) b[i] = 1;
  for (int(i) = 0; (i) < (int)(1000005); (i)++) nxt[i] = n + 1;
  init();
  scanf("%d", &m);
  for (int l, r, i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    ap[l].push_back({r, i});
  }
  for (int i = n; i >= 1; i--) {
    int tar = a[i];
    while (tar > 1) {
      int pri = p[tar], tmp = 1;
      while (tar % pri == 0) {
        tar /= pri;
        tmp *= pri;
      }
      update(i, (long long)tmp / pri * (pri - 1) % 1000000007);
      if (nxt[pri] <= n) {
        update(nxt[pri], (long long)pri * inv[pri - 1] % 1000000007);
      }
      nxt[pri] = i;
    }
    for (auto c : ap[i]) {
      ans[c.second] = query(c.first);
    }
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) printf("%d\n", ans[i]);
}
int main() {
  fmain();
  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>
using namespace std;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int p[1000005], inv[1000005];
long long fastPow(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ans;
}
void init() {
  for (int i = 2; i < 1000005; i++)
    if (!p[i]) {
      for (int j = i; j < 1000005; j += i) {
        p[j] = i;
      }
    }
  inv[0] = 1;
  for (int(i) = 1; (i) <= (int)(1000005 - 1); (i)++)
    inv[i] = fastPow(i, 1000000007 - 2);
}
int a[1000005];
int b[1000005], nxt[1000005];
void update(int x, long long val) {
  for (int i = x; i <= n; i += i & -i) {
    b[i] = (long long)b[i] * val % 1000000007;
  }
}
int query(int x) {
  int ret = 1;
  for (; x > 0; ret = (long long)ret * b[x] % 1000000007, x -= x & -x)
    ;
  return ret;
}
vector<pair<int, int> > ap[1000005];
int ans[1000005];
void fmain() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) b[i] = 1;
  for (int(i) = 0; (i) < (int)(1000005); (i)++) nxt[i] = n + 1;
  init();
  scanf("%d", &m);
  for (int l, r, i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    ap[l].push_back({r, i});
  }
  for (int i = n; i >= 1; i--) {
    int tar = a[i];
    while (tar > 1) {
      int pri = p[tar], tmp = 1;
      while (tar % pri == 0) {
        tar /= pri;
        tmp *= pri;
      }
      update(i, (long long)tmp / pri * (pri - 1) % 1000000007);
      if (nxt[pri] <= n) {
        update(nxt[pri], (long long)pri * inv[pri - 1] % 1000000007);
      }
      nxt[pri] = i;
    }
    for (auto c : ap[i]) {
      ans[c.second] = query(c.first);
    }
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) printf("%d\n", ans[i]);
}
int main() {
  fmain();
  return 0;
}

```