#include <bits/stdc++.h>
using namespace std;
const int MN = 2e5 + 200, MX = 1e6 + 100, mod = 1e9 + 7;
vector<int> primes[MX];
vector<int> qur[MN];
bool isp[MX];
int a[MN];
int rev[MX];
int l[MN], r[MN], ansq[MN];
int ls[MX];
int seg[4 * MN];
int n;
int powm(int x, int y) {
  if (y == 0) return 1;
  int ret = powm(x, y / 2);
  ret = (1ll * ret * ret) % mod;
  if (y & 1) ret = (1ll * ret * x) % mod;
  return ret;
}
void build(int v = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[v] = a[s];
    return;
  }
  int mid = (s + e) / 2;
  build(2 * v, s, mid);
  build(2 * v + 1, mid, e);
  seg[v] = (1ll * seg[2 * v] * seg[2 * v + 1]) % mod;
}
void upd(int p, int val, int v = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[v] = (1ll * seg[v] * val) % mod;
    return;
  }
  int mid = (s + e) / 2;
  if (p < mid) {
    upd(p, val, 2 * v, s, mid);
  } else {
    upd(p, val, 2 * v + 1, mid, e);
  }
  seg[v] = (1ll * seg[2 * v] * seg[2 * v + 1]) % mod;
}
int query(int l, int r, int v = 1, int s = 0, int e = n) {
  if (l <= s && e <= r) {
    return seg[v];
  } else if (l >= e || s >= r) {
    return 1;
  }
  int mid = (s + e) / 2;
  return (1ll * query(l, r, 2 * v, s, mid) * query(l, r, 2 * v + 1, mid, e)) %
         mod;
}
void add(int x) {
  for (int p : primes[a[x]]) {
    if (ls[p] != -1) {
      upd(ls[p], (1ll * p * rev[p - 1]) % mod);
    }
    ls[p] = x;
    upd(x, (1ll * (p - 1) * rev[p]) % mod);
  }
}
int main() {
  isp[1] = 1;
  for (int i = 1; i < MX; i++) {
    if (!isp[i]) {
      for (int j = i; j < MX; j += i) {
        isp[j] = 1;
        primes[j].push_back(i);
      }
    }
    rev[i] = powm(i, mod - 2);
    ls[i] = -1;
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &l[i], &r[i]);
    l[i]--;
    r[i]--;
    qur[r[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    add(i);
    for (int x : qur[i]) {
      ansq[x] = query(l[x], r[x] + 1);
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ansq[i]);
}

### Prompt

Please provide a Cpp coded solution 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;
const int MN = 2e5 + 200, MX = 1e6 + 100, mod = 1e9 + 7;
vector<int> primes[MX];
vector<int> qur[MN];
bool isp[MX];
int a[MN];
int rev[MX];
int l[MN], r[MN], ansq[MN];
int ls[MX];
int seg[4 * MN];
int n;
int powm(int x, int y) {
  if (y == 0) return 1;
  int ret = powm(x, y / 2);
  ret = (1ll * ret * ret) % mod;
  if (y & 1) ret = (1ll * ret * x) % mod;
  return ret;
}
void build(int v = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[v] = a[s];
    return;
  }
  int mid = (s + e) / 2;
  build(2 * v, s, mid);
  build(2 * v + 1, mid, e);
  seg[v] = (1ll * seg[2 * v] * seg[2 * v + 1]) % mod;
}
void upd(int p, int val, int v = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[v] = (1ll * seg[v] * val) % mod;
    return;
  }
  int mid = (s + e) / 2;
  if (p < mid) {
    upd(p, val, 2 * v, s, mid);
  } else {
    upd(p, val, 2 * v + 1, mid, e);
  }
  seg[v] = (1ll * seg[2 * v] * seg[2 * v + 1]) % mod;
}
int query(int l, int r, int v = 1, int s = 0, int e = n) {
  if (l <= s && e <= r) {
    return seg[v];
  } else if (l >= e || s >= r) {
    return 1;
  }
  int mid = (s + e) / 2;
  return (1ll * query(l, r, 2 * v, s, mid) * query(l, r, 2 * v + 1, mid, e)) %
         mod;
}
void add(int x) {
  for (int p : primes[a[x]]) {
    if (ls[p] != -1) {
      upd(ls[p], (1ll * p * rev[p - 1]) % mod);
    }
    ls[p] = x;
    upd(x, (1ll * (p - 1) * rev[p]) % mod);
  }
}
int main() {
  isp[1] = 1;
  for (int i = 1; i < MX; i++) {
    if (!isp[i]) {
      for (int j = i; j < MX; j += i) {
        isp[j] = 1;
        primes[j].push_back(i);
      }
    }
    rev[i] = powm(i, mod - 2);
    ls[i] = -1;
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &l[i], &r[i]);
    l[i]--;
    r[i]--;
    qur[r[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    add(i);
    for (int x : qur[i]) {
      ansq[x] = query(l[x], r[x] + 1);
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ansq[i]);
}

```