#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 7;
const long long LINF = (long long)1e18 + 9;
const int MXN = (int)1e6 + 10;
const double EPS = (double)1e-15;
int lp[MXN];
int n, m;
int a[MXN];
vector<pair<int, int> > queries[MXN];
int ans[MXN];
int t[MXN];
set<pair<int, int> > s;
void upd(int x, int y) {
  for (; x <= n; x += (x & (-x))) t[x] = (t[x] * 1LL * y) % INF;
}
int get(int x) {
  int res = 1;
  for (; x >= 1; x -= (x & (-x))) res = (res * 1LL * t[x]) % INF;
  return res;
}
int binpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (res * 1LL * x) % INF;
    x = (x * 1LL * x) % INF;
    y >>= 1;
  }
  return res;
}
void add(int pos, int x) {
  upd(pos, x);
  int tmp, tmp1;
  while (x > 1) {
    tmp = lp[x];
    auto it = s.lower_bound(make_pair(tmp, -1));
    if (it != s.end() && it->first == tmp) {
      tmp1 = (tmp * 1LL * binpow(tmp - 1, INF - 2)) % INF;
      upd(it->second, tmp1);
      s.erase(it);
    }
    tmp1 = ((tmp - 1) * 1LL * binpow(tmp, INF - 2)) % INF;
    upd(pos, tmp1);
    s.insert(make_pair(tmp, pos));
    while (x % tmp == 0) {
      x /= tmp;
    }
  }
}
int main() {
  for (int i = 0; i < MXN; ++i) t[i] = 1;
  for (int i = 2; i < MXN; ++i) {
    if (!lp[i]) {
      for (int j = i; j < MXN; j += i) {
        if (!lp[j]) lp[j] = i;
      }
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    queries[r].push_back(make_pair(l, i));
  }
  int res;
  for (int i = 1; i <= n; ++i) {
    add(i, a[i]);
    for (auto query : queries[i]) {
      res = (get(i) * 1LL * binpow(get(query.first - 1), INF - 2)) % INF;
      ans[query.second] = res;
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 INF = (int)1e9 + 7;
const long long LINF = (long long)1e18 + 9;
const int MXN = (int)1e6 + 10;
const double EPS = (double)1e-15;
int lp[MXN];
int n, m;
int a[MXN];
vector<pair<int, int> > queries[MXN];
int ans[MXN];
int t[MXN];
set<pair<int, int> > s;
void upd(int x, int y) {
  for (; x <= n; x += (x & (-x))) t[x] = (t[x] * 1LL * y) % INF;
}
int get(int x) {
  int res = 1;
  for (; x >= 1; x -= (x & (-x))) res = (res * 1LL * t[x]) % INF;
  return res;
}
int binpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (res * 1LL * x) % INF;
    x = (x * 1LL * x) % INF;
    y >>= 1;
  }
  return res;
}
void add(int pos, int x) {
  upd(pos, x);
  int tmp, tmp1;
  while (x > 1) {
    tmp = lp[x];
    auto it = s.lower_bound(make_pair(tmp, -1));
    if (it != s.end() && it->first == tmp) {
      tmp1 = (tmp * 1LL * binpow(tmp - 1, INF - 2)) % INF;
      upd(it->second, tmp1);
      s.erase(it);
    }
    tmp1 = ((tmp - 1) * 1LL * binpow(tmp, INF - 2)) % INF;
    upd(pos, tmp1);
    s.insert(make_pair(tmp, pos));
    while (x % tmp == 0) {
      x /= tmp;
    }
  }
}
int main() {
  for (int i = 0; i < MXN; ++i) t[i] = 1;
  for (int i = 2; i < MXN; ++i) {
    if (!lp[i]) {
      for (int j = i; j < MXN; j += i) {
        if (!lp[j]) lp[j] = i;
      }
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    queries[r].push_back(make_pair(l, i));
  }
  int res;
  for (int i = 1; i <= n; ++i) {
    add(i, a[i]);
    for (auto query : queries[i]) {
      res = (get(i) * 1LL * binpow(get(query.first - 1), INF - 2)) % INF;
      ans[query.second] = res;
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```