#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e6 + 5, mo = 1e9 + 7;
int n, m, Q, i, a[N], p[M], ans[N];
int he[N];
struct edge {
  int l, to;
} e[N];
int read() {
  char c = getchar();
  int k = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) k = (k << 3) + (k << 1) + c - 48;
  return k;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
int t, pr[M], d[M], inv[M];
void init(int n) {
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!d[i]) pr[++t] = i, d[i] = i;
    for (int j = 1; i * pr[j] <= n; j++) {
      d[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) break;
    }
    inv[i] = (long long)inv[mo % i] * (mo - mo / i) % mo;
  }
}
int f[N];
void add(int x, int k) {
  for (; x <= n; x += x & -x) f[x] = (long long)f[x] * k % mo;
}
int ask(int x) {
  int k = 1;
  for (; x; x -= x & -x) k = (long long)k * f[x] % mo;
  return k;
}
int main() {
  for (n = read(), i = 1; i <= n; i++) m = max(m, a[i] = read());
  init(m);
  for (Q = read(), i = 1; i <= Q; i++) {
    int x = read(), y = read();
    e[i].l = he[x];
    he[x] = i;
    e[i].to = y;
  }
  for (i = 1; i <= n; i++) f[i] = 1;
  for (int x = n; x; x--) {
    add(x, a[x]);
    for (int k = a[x]; k > 1; k /= d[k]) {
      int dx = d[k];
      if (p[dx] == x) continue;
      int vx = (long long)(dx - 1) * inv[dx] % mo;
      int vy = (long long)dx * inv[dx - 1] % mo;
      if (p[dx]) add(p[dx], vy);
      p[dx] = x;
      add(x, vx);
    }
    for (i = he[x]; i; i = e[i].l) ans[i] = ask(e[i].to);
  }
  for (i = 1; i <= Q; i++) write(ans[i]), putchar('\n');
}

### Prompt

Create a solution in CPP for 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 int N = 2e5 + 5, M = 1e6 + 5, mo = 1e9 + 7;
int n, m, Q, i, a[N], p[M], ans[N];
int he[N];
struct edge {
  int l, to;
} e[N];
int read() {
  char c = getchar();
  int k = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) k = (k << 3) + (k << 1) + c - 48;
  return k;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
int t, pr[M], d[M], inv[M];
void init(int n) {
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!d[i]) pr[++t] = i, d[i] = i;
    for (int j = 1; i * pr[j] <= n; j++) {
      d[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) break;
    }
    inv[i] = (long long)inv[mo % i] * (mo - mo / i) % mo;
  }
}
int f[N];
void add(int x, int k) {
  for (; x <= n; x += x & -x) f[x] = (long long)f[x] * k % mo;
}
int ask(int x) {
  int k = 1;
  for (; x; x -= x & -x) k = (long long)k * f[x] % mo;
  return k;
}
int main() {
  for (n = read(), i = 1; i <= n; i++) m = max(m, a[i] = read());
  init(m);
  for (Q = read(), i = 1; i <= Q; i++) {
    int x = read(), y = read();
    e[i].l = he[x];
    he[x] = i;
    e[i].to = y;
  }
  for (i = 1; i <= n; i++) f[i] = 1;
  for (int x = n; x; x--) {
    add(x, a[x]);
    for (int k = a[x]; k > 1; k /= d[k]) {
      int dx = d[k];
      if (p[dx] == x) continue;
      int vx = (long long)(dx - 1) * inv[dx] % mo;
      int vy = (long long)dx * inv[dx - 1] % mo;
      if (p[dx]) add(p[dx], vy);
      p[dx] = x;
      add(x, vx);
    }
    for (i = he[x]; i; i = e[i].l) ans[i] = ask(e[i].to);
  }
  for (i = 1; i <= Q; i++) write(ans[i]), putchar('\n');
}

```