#include <bits/stdc++.h>
using namespace std;
const int N = 200 * 1000 + 5, M = 1000 * 1000 + 5, mod = 1000 * 1000 * 1000 + 7;
int n, q;
int a[N];
int last[N];
vector<int> d[M];
long long fi[M];
long long seg[N * 4];
vector<pair<int, int> > t[N];
long long ans[N];
int lt[M];
int po(int x, int y) {
  if (y == 0) {
    return 1;
  }
  long long ans = po(x, y / 2);
  ans *= ans;
  ans %= mod;
  if (y % 2) {
    ans *= x;
    ans %= mod;
  }
  return ans;
}
void update(int p, int val, int id = 1, int s = 0, int e = n) {
  if (e <= p || p < s) {
    return;
  }
  if (e - s == 1) {
    seg[id] *= val;
    seg[id] %= mod;
    return;
  }
  int md = (s + e) / 2;
  update(p, val, id * 2, s, md);
  update(p, val, id * 2 + 1, md, e);
  seg[id] = 1ll * seg[id * 2] * seg[id * 2 + 1] % mod;
}
int get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) {
    return 1;
  }
  if (l <= s && e <= r) {
    return seg[id];
  }
  int md = (s + e) / 2;
  long long ans = 1ll * get(l, r, id * 2, s, md) * get(l, r, id * 2 + 1, md, e);
  ans %= mod;
  return ans;
}
int main() {
  ios ::sync_with_stdio(false);
  iota(fi + 1, fi + M, 1);
  for (int i = 2; i < M; i++) {
    if (fi[i] == i) {
      for (int j = i; j < M; j += i) {
        d[j].push_back(i);
        fi[j] = fi[j] * (i - 1) / i;
      }
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  fill(seg, seg + n * 4, 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    t[r].push_back({l, i});
  }
  fill(lt, lt + M, -1);
  for (int i = 0; i < n; i++) {
    for (auto x : d[a[i]]) {
      if (lt[x] != -1) {
        long long k = 1ll * x * po(x - 1, mod - 2) % mod;
        update(lt[x], k);
      }
      lt[x] = i;
    }
    update(i, fi[a[i]]);
    for (auto p : t[i]) {
      ans[p.second] = get(p.first, i + 1);
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
  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 = 200 * 1000 + 5, M = 1000 * 1000 + 5, mod = 1000 * 1000 * 1000 + 7;
int n, q;
int a[N];
int last[N];
vector<int> d[M];
long long fi[M];
long long seg[N * 4];
vector<pair<int, int> > t[N];
long long ans[N];
int lt[M];
int po(int x, int y) {
  if (y == 0) {
    return 1;
  }
  long long ans = po(x, y / 2);
  ans *= ans;
  ans %= mod;
  if (y % 2) {
    ans *= x;
    ans %= mod;
  }
  return ans;
}
void update(int p, int val, int id = 1, int s = 0, int e = n) {
  if (e <= p || p < s) {
    return;
  }
  if (e - s == 1) {
    seg[id] *= val;
    seg[id] %= mod;
    return;
  }
  int md = (s + e) / 2;
  update(p, val, id * 2, s, md);
  update(p, val, id * 2 + 1, md, e);
  seg[id] = 1ll * seg[id * 2] * seg[id * 2 + 1] % mod;
}
int get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) {
    return 1;
  }
  if (l <= s && e <= r) {
    return seg[id];
  }
  int md = (s + e) / 2;
  long long ans = 1ll * get(l, r, id * 2, s, md) * get(l, r, id * 2 + 1, md, e);
  ans %= mod;
  return ans;
}
int main() {
  ios ::sync_with_stdio(false);
  iota(fi + 1, fi + M, 1);
  for (int i = 2; i < M; i++) {
    if (fi[i] == i) {
      for (int j = i; j < M; j += i) {
        d[j].push_back(i);
        fi[j] = fi[j] * (i - 1) / i;
      }
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  fill(seg, seg + n * 4, 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    t[r].push_back({l, i});
  }
  fill(lt, lt + M, -1);
  for (int i = 0; i < n; i++) {
    for (auto x : d[a[i]]) {
      if (lt[x] != -1) {
        long long k = 1ll * x * po(x - 1, mod - 2) % mod;
        update(lt[x], k);
      }
      lt[x] = i;
    }
    update(i, fi[a[i]]);
    for (auto p : t[i]) {
      ans[p.second] = get(p.first, i + 1);
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}

```