#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7;
vector<int> contain[M], nex[M], P;
int a[N], st[M], n, ind[N], ans[N];
pair<int, int> query[N];
long long tree[N];
void change(int x, int c) {
  if (x == 0) return;
  for (; x <= n; x += x & -x) (tree[x] *= c) %= MOD;
}
long long ask(int x) {
  long long res = 1;
  for (; x; x -= x & -x) (res *= tree[x]) %= MOD;
  return res;
}
bool cmp(int i, int j) { return query[i].second < query[j].second; }
long long mpow(long long x, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) (res *= x) %= MOD;
    x = (x * x) % MOD;
    n >>= 1;
  }
  return res;
}
void work(int x) {
  change(x, a[x]);
  for (int i = 0; i < (int)contain[a[x]].size(); ++i) {
    int pp = contain[a[x]][i];
    change(nex[x][i], pp);
    change(nex[x][i], mpow(pp - 1, MOD - 2));
    change(x, mpow(pp, MOD - 2));
    change(x, pp - 1);
  }
}
int main() {
  for (int i = 2; i < M; ++i) {
    if (contain[i].size() == 0) {
      for (int j = i; j < M; j += i) contain[j].push_back(i);
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (int)contain[a[i]].size(); ++j) {
      nex[i].push_back(st[contain[a[i]][j]]);
      st[contain[a[i]][j]] = i;
    }
  }
  for (int i = 1; i <= n; ++i) tree[i] = 1;
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &query[i].first, &query[i].second);
    ind[i] = i;
  }
  sort(ind + 1, ind + 1 + m, cmp);
  for (int i = 1, now = 0; i <= m; ++i) {
    int L = query[ind[i]].first, R = query[ind[i]].second;
    while (now < R) work(++now);
    ans[ind[i]] = (ask(R) * mpow(ask(L - 1), MOD - 2)) % MOD;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 2e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7;
vector<int> contain[M], nex[M], P;
int a[N], st[M], n, ind[N], ans[N];
pair<int, int> query[N];
long long tree[N];
void change(int x, int c) {
  if (x == 0) return;
  for (; x <= n; x += x & -x) (tree[x] *= c) %= MOD;
}
long long ask(int x) {
  long long res = 1;
  for (; x; x -= x & -x) (res *= tree[x]) %= MOD;
  return res;
}
bool cmp(int i, int j) { return query[i].second < query[j].second; }
long long mpow(long long x, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) (res *= x) %= MOD;
    x = (x * x) % MOD;
    n >>= 1;
  }
  return res;
}
void work(int x) {
  change(x, a[x]);
  for (int i = 0; i < (int)contain[a[x]].size(); ++i) {
    int pp = contain[a[x]][i];
    change(nex[x][i], pp);
    change(nex[x][i], mpow(pp - 1, MOD - 2));
    change(x, mpow(pp, MOD - 2));
    change(x, pp - 1);
  }
}
int main() {
  for (int i = 2; i < M; ++i) {
    if (contain[i].size() == 0) {
      for (int j = i; j < M; j += i) contain[j].push_back(i);
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (int)contain[a[i]].size(); ++j) {
      nex[i].push_back(st[contain[a[i]][j]]);
      st[contain[a[i]][j]] = i;
    }
  }
  for (int i = 1; i <= n; ++i) tree[i] = 1;
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &query[i].first, &query[i].second);
    ind[i] = i;
  }
  sort(ind + 1, ind + 1 + m, cmp);
  for (int i = 1, now = 0; i <= m; ++i) {
    int L = query[ind[i]].first, R = query[ind[i]].second;
    while (now < R) work(++now);
    ans[ind[i]] = (ask(R) * mpow(ask(L - 1), MOD - 2)) % MOD;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```