#include <bits/stdc++.h>
using namespace std;
vector<int> pos[1001000];
vector<int> pf[1001000];
int lpf[1001000], prm[1001000], inv[1001000], pcnt;
int ans[200100], a[200100], f[200100];
bool notp[1001000];
struct qry {
  int l, r, id;
  bool operator<(const qry& o) const { return l < o.l; }
} qq[200100];
int cal_inv(int x) {
  int y = 1000000007 - 2, r = 1;
  while (y) {
    if (y & 1) r = 1LL * r * x % 1000000007;
    x = 1LL * x * x % 1000000007;
    y >>= 1;
  }
  return r;
}
void upd(int x, int v) {
  while (x < 200100) {
    f[x] = 1LL * v * f[x] % 1000000007;
    x += x & -x;
  }
}
int mul(int x) {
  int r = 1;
  while (x) {
    r = 1LL * r * f[x] % 1000000007;
    x -= x & -x;
  }
  return r;
}
int main() {
  for (int i = 2; i < 1001000; i++) {
    if (!notp[i]) prm[pcnt++] = i, lpf[i] = i;
    for (int j = 0; j < pcnt && i * prm[j] < 1001000; j++) {
      notp[i * prm[j]] = true;
      lpf[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
  for (int i = 1; i < 1001000; i++) {
    inv[i] = cal_inv(i);
  }
  for (int i = 1; i < 200100; i++) f[i] = 1;
  int n, q, now = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    upd(i, a[i]);
  }
  for (int i = n; i >= 1; i--) {
    while (a[i] > 1) {
      int p = lpf[a[i]];
      pf[i].push_back(p);
      pos[p].push_back(i);
      while (a[i] % p == 0) a[i] /= p;
    }
  }
  for (int i = 0; i < pcnt; i++) {
    if (!pos[prm[i]].empty()) {
      upd(pos[prm[i]].back(), 1LL * inv[prm[i]] * (prm[i] - 1) % 1000000007);
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &qq[i].l, &qq[i].r);
    qq[i].id = i;
  }
  sort(qq, qq + q);
  for (int i = 0; i < q; i++) {
    while (now < qq[i].l) {
      upd(now, inv[f[now]]);
      for (int j = 0; j < pf[now].size(); j++) {
        int p = pf[now][j];
        pos[p].pop_back();
        if (!pos[p].empty()) {
          upd(pos[p].back(), 1LL * inv[p] * (p - 1) % 1000000007);
        }
      }
      now++;
    }
    ans[qq[i].id] = mul(qq[i].r);
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Generate 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;
vector<int> pos[1001000];
vector<int> pf[1001000];
int lpf[1001000], prm[1001000], inv[1001000], pcnt;
int ans[200100], a[200100], f[200100];
bool notp[1001000];
struct qry {
  int l, r, id;
  bool operator<(const qry& o) const { return l < o.l; }
} qq[200100];
int cal_inv(int x) {
  int y = 1000000007 - 2, r = 1;
  while (y) {
    if (y & 1) r = 1LL * r * x % 1000000007;
    x = 1LL * x * x % 1000000007;
    y >>= 1;
  }
  return r;
}
void upd(int x, int v) {
  while (x < 200100) {
    f[x] = 1LL * v * f[x] % 1000000007;
    x += x & -x;
  }
}
int mul(int x) {
  int r = 1;
  while (x) {
    r = 1LL * r * f[x] % 1000000007;
    x -= x & -x;
  }
  return r;
}
int main() {
  for (int i = 2; i < 1001000; i++) {
    if (!notp[i]) prm[pcnt++] = i, lpf[i] = i;
    for (int j = 0; j < pcnt && i * prm[j] < 1001000; j++) {
      notp[i * prm[j]] = true;
      lpf[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
  for (int i = 1; i < 1001000; i++) {
    inv[i] = cal_inv(i);
  }
  for (int i = 1; i < 200100; i++) f[i] = 1;
  int n, q, now = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    upd(i, a[i]);
  }
  for (int i = n; i >= 1; i--) {
    while (a[i] > 1) {
      int p = lpf[a[i]];
      pf[i].push_back(p);
      pos[p].push_back(i);
      while (a[i] % p == 0) a[i] /= p;
    }
  }
  for (int i = 0; i < pcnt; i++) {
    if (!pos[prm[i]].empty()) {
      upd(pos[prm[i]].back(), 1LL * inv[prm[i]] * (prm[i] - 1) % 1000000007);
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &qq[i].l, &qq[i].r);
    qq[i].id = i;
  }
  sort(qq, qq + q);
  for (int i = 0; i < q; i++) {
    while (now < qq[i].l) {
      upd(now, inv[f[now]]);
      for (int j = 0; j < pf[now].size(); j++) {
        int p = pf[now][j];
        pos[p].pop_back();
        if (!pos[p].empty()) {
          upd(pos[p].back(), 1LL * inv[p] * (p - 1) % 1000000007);
        }
      }
      now++;
    }
    ans[qq[i].id] = mul(qq[i].r);
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}

```