#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long fwt[200005], p[200005], arr[200005], ans[200005], lp[1000005],
    b[1000005];
int n, m;
struct node {
  int l, r, id;
} q[200005];
bool cmp(node a, node b) { return a.r < b.r; }
long long inv(long long x) {
  long long e = mod - 2, rs = 1;
  for (; e; e >>= 1) {
    if (e & 1) rs = rs * 1ll * x % mod;
    x = x * 1ll * x % mod;
  }
  return rs;
}
void upd(int idx, long long val) {
  for (; idx <= n; idx += idx & -idx) fwt[idx] = fwt[idx] * 1ll * val % mod;
}
long long que(int idx) {
  long long rs = 1ll;
  for (; idx; idx -= idx & -idx) rs = rs * 1ll * fwt[idx] % mod;
  return rs;
}
void change(int p) {
  int x = arr[p];
  while (x != 1) {
    int nw = lp[x];
    if (b[nw]) {
      upd(b[nw], inv(nw - 1));
      upd(b[nw], nw);
    }
    b[nw] = p;
    upd(p, nw - 1);
    x /= nw;
  }
}
int main() {
  scanf("%d", &n);
  long long mx = -1;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i]);
    mx = max(mx, arr[i]);
    fwt[i] = 1ll;
  }
  fwt[0] = 1ll;
  scanf("%d", &m);
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  int cnt = 0;
  for (long long i = 2; i <= mx; i++) {
    if (!lp[i]) {
      lp[i] = i;
      p[++cnt] = i;
    }
    for (int j = 1; j <= cnt; j++) {
      if (p[j] * i > mx) break;
      lp[p[j] * i] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1, nw = 1; i <= m; i++) {
    while (nw <= q[i].r) change(nw++);
    ans[q[i].id] = que(q[i].r) * 1ll * inv(que(q[i].l - 1)) % mod;
  }
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}

### Prompt

Please formulate 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;
const long long mod = 1e9 + 7;
long long fwt[200005], p[200005], arr[200005], ans[200005], lp[1000005],
    b[1000005];
int n, m;
struct node {
  int l, r, id;
} q[200005];
bool cmp(node a, node b) { return a.r < b.r; }
long long inv(long long x) {
  long long e = mod - 2, rs = 1;
  for (; e; e >>= 1) {
    if (e & 1) rs = rs * 1ll * x % mod;
    x = x * 1ll * x % mod;
  }
  return rs;
}
void upd(int idx, long long val) {
  for (; idx <= n; idx += idx & -idx) fwt[idx] = fwt[idx] * 1ll * val % mod;
}
long long que(int idx) {
  long long rs = 1ll;
  for (; idx; idx -= idx & -idx) rs = rs * 1ll * fwt[idx] % mod;
  return rs;
}
void change(int p) {
  int x = arr[p];
  while (x != 1) {
    int nw = lp[x];
    if (b[nw]) {
      upd(b[nw], inv(nw - 1));
      upd(b[nw], nw);
    }
    b[nw] = p;
    upd(p, nw - 1);
    x /= nw;
  }
}
int main() {
  scanf("%d", &n);
  long long mx = -1;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i]);
    mx = max(mx, arr[i]);
    fwt[i] = 1ll;
  }
  fwt[0] = 1ll;
  scanf("%d", &m);
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  int cnt = 0;
  for (long long i = 2; i <= mx; i++) {
    if (!lp[i]) {
      lp[i] = i;
      p[++cnt] = i;
    }
    for (int j = 1; j <= cnt; j++) {
      if (p[j] * i > mx) break;
      lp[p[j] * i] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1, nw = 1; i <= m; i++) {
    while (nw <= q[i].r) change(nw++);
    ans[q[i].id] = que(q[i].r) * 1ll * inv(que(q[i].l - 1)) % mod;
  }
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}

```