#include <bits/stdc++.h>
int n, a[200011], minp[1000011], b[1000011], v, prime[1000011], t[200011], m,
    ans[200011];
const long long mod = 1e9 + 7;
using namespace std;
struct Node {
  int l, r, num;
} q[200011];
inline bool cmp(Node a, Node b) { return a.r < b.r; }
inline void predeal() {
  int tot = 0;
  for (int i = 2; i <= v; i++) {
    if (!minp[i]) prime[tot++] = i, minp[i] = i;
    for (int j = 0; j < tot; j++) {
      if (prime[j] * i > v) break;
      minp[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 0; i <= n; i++) t[i] = 1;
}
inline int pow(int a, long long b) {
  int r = 1, base = a;
  while (b) {
    if (b & 1) r = (long long)r * base % mod;
    base = (long long)base * base % mod;
    b >>= 1;
  }
  return r % mod;
}
inline void modify(int x, int z) {
  for (int i = x; i <= n; i += i & -i) t[i] = (long long)t[i] * z % mod;
}
inline int query(int x) {
  int temp = 1;
  for (int i = x; i; i -= i & -i) temp = (long long)temp * t[i] % mod;
  return temp;
}
inline void change(int p) {
  int x = a[p];
  while (x != 1) {
    int now = minp[x];
    if (b[now]) {
      modify(b[now], pow(now - 1, mod - 2));
      modify(b[now], now);
    }
    b[now] = p;
    modify(p, now - 1);
    x /= now;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  v = *max_element(a + 1, a + 1 + n);
  predeal();
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].num = i;
  }
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1, now = 1; i <= m; i++) {
    while (now <= q[i].r) change(now++);
    ans[q[i].num] =
        (long long)query(q[i].r) * pow(query(q[i].l - 1), mod - 2) % mod;
  }
  for (int i = 1; i <= m; 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>
int n, a[200011], minp[1000011], b[1000011], v, prime[1000011], t[200011], m,
    ans[200011];
const long long mod = 1e9 + 7;
using namespace std;
struct Node {
  int l, r, num;
} q[200011];
inline bool cmp(Node a, Node b) { return a.r < b.r; }
inline void predeal() {
  int tot = 0;
  for (int i = 2; i <= v; i++) {
    if (!minp[i]) prime[tot++] = i, minp[i] = i;
    for (int j = 0; j < tot; j++) {
      if (prime[j] * i > v) break;
      minp[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 0; i <= n; i++) t[i] = 1;
}
inline int pow(int a, long long b) {
  int r = 1, base = a;
  while (b) {
    if (b & 1) r = (long long)r * base % mod;
    base = (long long)base * base % mod;
    b >>= 1;
  }
  return r % mod;
}
inline void modify(int x, int z) {
  for (int i = x; i <= n; i += i & -i) t[i] = (long long)t[i] * z % mod;
}
inline int query(int x) {
  int temp = 1;
  for (int i = x; i; i -= i & -i) temp = (long long)temp * t[i] % mod;
  return temp;
}
inline void change(int p) {
  int x = a[p];
  while (x != 1) {
    int now = minp[x];
    if (b[now]) {
      modify(b[now], pow(now - 1, mod - 2));
      modify(b[now], now);
    }
    b[now] = p;
    modify(p, now - 1);
    x /= now;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  v = *max_element(a + 1, a + 1 + n);
  predeal();
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].num = i;
  }
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1, now = 1; i <= m; i++) {
    while (now <= q[i].r) change(now++);
    ans[q[i].num] =
        (long long)query(q[i].r) * pow(query(q[i].l - 1), mod - 2) % mod;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```