#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, MAXN = 1e6 + 10, mod = 1e9 + 7;
vector<int> pr[MAXN];
vector<pair<int, int> > q[N];
bool mark[MAXN];
int n, t, val, a[N], seg[N << 2], lst[MAXN], pm[N], ans[N];
int power(int a, int b) {
  int rtr = 1;
  for (; b > 0; b /= 2, a = 1ll * a * a % mod)
    if (b & 1) rtr = 1ll * rtr * a % mod;
  return rtr;
}
void Er() {
  for (int i = 2; i < MAXN - 5; i++) {
    if (!mark[i]) {
      for (int j = i; j < MAXN - 5; j += i) mark[j] = true, pr[j].push_back(i);
    }
  }
  return;
}
int Prd(int l, int r) { return 1ll * pm[r] * power(pm[l], mod - 2) % mod; }
void upd(int p, int id = 1, int s = 0, int e = n) {
  seg[id] = 1ll * seg[id] * val % mod;
  if (e - s == 1) return;
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  if (p < mid)
    upd(p, lc, s, mid);
  else
    upd(p, rc, mid, e);
  return;
}
int get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (e <= l || r <= s) return 1;
  if (l <= s && e <= r) return seg[id];
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  return 1ll * get(l, r, lc, s, mid) * get(l, r, rc, mid, e) % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  Er();
  memset(lst, -1, sizeof(lst));
  pm[0] = 1;
  cin >> n;
  for (int i = 0; i <= 4 * n; i++) seg[i] = 1;
  for (int i = 0; i < n; i++) cin >> a[i], pm[i + 1] = 1ll * pm[i] * a[i] % mod;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    q[r].push_back({l, i});
  }
  for (int i = 0; i < n; i++) {
    for (auto u : pr[a[i]]) {
      val = (1 - power(u, mod - 2) + mod) % mod, upd(i);
      if (lst[u] != -1) val = power(val, mod - 2), upd(lst[u]);
      lst[u] = i;
    }
    for (auto p : q[i + 1]) {
      int rn = get(p.first, i + 1);
      ans[p.second] = 1ll * Prd(p.first, i + 1) * rn % mod;
    }
  }
  for (int i = 0; i < t; 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 = 2e5 + 10, MAXN = 1e6 + 10, mod = 1e9 + 7;
vector<int> pr[MAXN];
vector<pair<int, int> > q[N];
bool mark[MAXN];
int n, t, val, a[N], seg[N << 2], lst[MAXN], pm[N], ans[N];
int power(int a, int b) {
  int rtr = 1;
  for (; b > 0; b /= 2, a = 1ll * a * a % mod)
    if (b & 1) rtr = 1ll * rtr * a % mod;
  return rtr;
}
void Er() {
  for (int i = 2; i < MAXN - 5; i++) {
    if (!mark[i]) {
      for (int j = i; j < MAXN - 5; j += i) mark[j] = true, pr[j].push_back(i);
    }
  }
  return;
}
int Prd(int l, int r) { return 1ll * pm[r] * power(pm[l], mod - 2) % mod; }
void upd(int p, int id = 1, int s = 0, int e = n) {
  seg[id] = 1ll * seg[id] * val % mod;
  if (e - s == 1) return;
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  if (p < mid)
    upd(p, lc, s, mid);
  else
    upd(p, rc, mid, e);
  return;
}
int get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (e <= l || r <= s) return 1;
  if (l <= s && e <= r) return seg[id];
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  return 1ll * get(l, r, lc, s, mid) * get(l, r, rc, mid, e) % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  Er();
  memset(lst, -1, sizeof(lst));
  pm[0] = 1;
  cin >> n;
  for (int i = 0; i <= 4 * n; i++) seg[i] = 1;
  for (int i = 0; i < n; i++) cin >> a[i], pm[i + 1] = 1ll * pm[i] * a[i] % mod;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    q[r].push_back({l, i});
  }
  for (int i = 0; i < n; i++) {
    for (auto u : pr[a[i]]) {
      val = (1 - power(u, mod - 2) + mod) % mod, upd(i);
      if (lst[u] != -1) val = power(val, mod - 2), upd(lst[u]);
      lst[u] = i;
    }
    for (auto p : q[i + 1]) {
      int rn = get(p.first, i + 1);
      ans[p.second] = 1ll * Prd(p.first, i + 1) * rn % mod;
    }
  }
  for (int i = 0; i < t; i++) cout << ans[i] << "\n";
  return 0;
}

```