#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e6 + 5, mod = 1e9 + 7;
int bpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    b >>= 1;
    a = 1ll * a * a % mod;
  }
  return ans;
}
int fen[N];
void mul(int p, int val) {
  for (p++; p < N; p += p & -p) fen[p] = 1LL * fen[p] * val % mod;
}
int _get(int p) {
  int ans = 1;
  for (; p > 0; p -= p & -p) ans = 1LL * ans * fen[p] % mod;
  return ans;
}
int get(int l, int r) { return 1LL * _get(r) * bpow(_get(l), mod - 2) % mod; }
int last[M], a[N], ans[N], rev[M];
vector<int> ps[M];
vector<pair<int, int> > qs[N];
void test() {
  while (1) {
    int n;
    cin >> n;
    for (int i = 0; i < ps[n].size(); i++) cout << ps[n][i] << " ";
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  for (int i = 0; i < N; i++) fen[i] = 1;
  for (int i = 2; i < M; i++) {
    if (ps[i].size()) continue;
    for (int j = i; j < M; j += i) ps[j].push_back(i);
  }
  for (int i = 0; i < M; i++) rev[i] = bpow(i, mod - 2);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    qs[r].push_back(make_pair(l, i));
  }
  memset(last, -1, sizeof last);
  for (int i = 0; i < n; i++) {
    mul(i, a[i]);
    for (int j = 0; j < ps[a[i]].size(); j++) {
      int np = ps[a[i]][j];
      int fp = 1ll * (np - 1) * rev[np] % mod;
      if (last[np] != -1) mul(last[np], 1ll * np * rev[np - 1] % mod);
      last[np] = i;
      mul(i, fp);
    }
    for (int j = 0; j < qs[i].size(); j++) {
      pair<int, int> q = qs[i][j];
      ans[q.second] = get(q.first, i + 1);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  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 N = 2e5 + 5, M = 1e6 + 5, mod = 1e9 + 7;
int bpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    b >>= 1;
    a = 1ll * a * a % mod;
  }
  return ans;
}
int fen[N];
void mul(int p, int val) {
  for (p++; p < N; p += p & -p) fen[p] = 1LL * fen[p] * val % mod;
}
int _get(int p) {
  int ans = 1;
  for (; p > 0; p -= p & -p) ans = 1LL * ans * fen[p] % mod;
  return ans;
}
int get(int l, int r) { return 1LL * _get(r) * bpow(_get(l), mod - 2) % mod; }
int last[M], a[N], ans[N], rev[M];
vector<int> ps[M];
vector<pair<int, int> > qs[N];
void test() {
  while (1) {
    int n;
    cin >> n;
    for (int i = 0; i < ps[n].size(); i++) cout << ps[n][i] << " ";
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  for (int i = 0; i < N; i++) fen[i] = 1;
  for (int i = 2; i < M; i++) {
    if (ps[i].size()) continue;
    for (int j = i; j < M; j += i) ps[j].push_back(i);
  }
  for (int i = 0; i < M; i++) rev[i] = bpow(i, mod - 2);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    qs[r].push_back(make_pair(l, i));
  }
  memset(last, -1, sizeof last);
  for (int i = 0; i < n; i++) {
    mul(i, a[i]);
    for (int j = 0; j < ps[a[i]].size(); j++) {
      int np = ps[a[i]][j];
      int fp = 1ll * (np - 1) * rev[np] % mod;
      if (last[np] != -1) mul(last[np], 1ll * np * rev[np - 1] % mod);
      last[np] = i;
      mul(i, fp);
    }
    for (int j = 0; j < qs[i].size(); j++) {
      pair<int, int> q = qs[i][j];
      ans[q.second] = get(q.first, i + 1);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}

```