#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1100000;
int prime[N], num_prime, first[N], num, n, a[N], rightmost[N], ni[N], ans[N];
long long pi[N];
struct ones {
  int l, r, id;
  ones() {}
  ones(int l, int r, int id) : l(l), r(r), id(id) {}
} f[N * 2];
long long tree[N];
long long mi(long long x, long long n) {
  if (n == 0) return 1;
  long long tmp = mi(x, n / 2);
  tmp = (tmp * tmp) % MOD;
  if (n % 2 == 1) tmp = (tmp * x) % MOD;
  return tmp;
}
int lowbit(int x) { return x & (-x); }
void change(int x, long long delt) {
  for (int i = x; i > 0; i -= lowbit(i)) tree[i] = tree[i] * delt % MOD;
}
long long get(int x) {
  long long s = 1;
  for (int i = x; i < N; i += lowbit(i)) s = s * tree[i] % MOD;
  return s;
}
void add(int t) {
  int x = a[t];
  while (x != 1) {
    int k = first[x];
    change(rightmost[k] - 1,
           (long long)(prime[k] - 1) * mi(prime[k], MOD - 2) % MOD);
    change(t - 1, mi(prime[k] - 1, MOD - 2) * (long long)(prime[k]) % MOD);
    rightmost[k] = t;
    while (x % prime[k] == 0) x /= prime[k];
  }
}
bool cmp(const ones &a, const ones &b) { return a.l > b.l; }
int main() {
  cin >> n;
  num_prime = 0;
  for (int i = 2; i <= N; i++) {
    if (!first[i]) {
      prime[++num_prime] = i;
      first[i] = num_prime;
    }
    for (int j = 1; j <= num_prime && i * prime[j] < N; j++) {
      first[prime[j] * i] = j;
      if (i % prime[j] == 0) break;
    }
  }
  pi[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pi[i] = pi[i - 1] * (long long)a[i] % MOD;
  }
  for (int i = 1; i < N; i++) rightmost[i] = N - 10;
  int num_q;
  cin >> num_q;
  num = 0;
  for (int i = 1; i <= num_q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    f[++num] = ones(l, r, i);
  }
  sort(f + 1, f + num + 1, cmp);
  for (int i = 0; i < N; i++) tree[i] = 1;
  int zz = n;
  for (int i = 1; i <= num; i++) {
    while (zz >= f[i].l) {
      add(zz);
      zz--;
    }
    long long tmp = pi[f[i].r] * mi(pi[f[i].l - 1], MOD - 2) % MOD;
    tmp = tmp * get(f[i].r) % MOD;
    ans[f[i].id] = tmp;
  }
  for (int i = 1; i <= num_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;
const int MOD = 1e9 + 7;
const int N = 1100000;
int prime[N], num_prime, first[N], num, n, a[N], rightmost[N], ni[N], ans[N];
long long pi[N];
struct ones {
  int l, r, id;
  ones() {}
  ones(int l, int r, int id) : l(l), r(r), id(id) {}
} f[N * 2];
long long tree[N];
long long mi(long long x, long long n) {
  if (n == 0) return 1;
  long long tmp = mi(x, n / 2);
  tmp = (tmp * tmp) % MOD;
  if (n % 2 == 1) tmp = (tmp * x) % MOD;
  return tmp;
}
int lowbit(int x) { return x & (-x); }
void change(int x, long long delt) {
  for (int i = x; i > 0; i -= lowbit(i)) tree[i] = tree[i] * delt % MOD;
}
long long get(int x) {
  long long s = 1;
  for (int i = x; i < N; i += lowbit(i)) s = s * tree[i] % MOD;
  return s;
}
void add(int t) {
  int x = a[t];
  while (x != 1) {
    int k = first[x];
    change(rightmost[k] - 1,
           (long long)(prime[k] - 1) * mi(prime[k], MOD - 2) % MOD);
    change(t - 1, mi(prime[k] - 1, MOD - 2) * (long long)(prime[k]) % MOD);
    rightmost[k] = t;
    while (x % prime[k] == 0) x /= prime[k];
  }
}
bool cmp(const ones &a, const ones &b) { return a.l > b.l; }
int main() {
  cin >> n;
  num_prime = 0;
  for (int i = 2; i <= N; i++) {
    if (!first[i]) {
      prime[++num_prime] = i;
      first[i] = num_prime;
    }
    for (int j = 1; j <= num_prime && i * prime[j] < N; j++) {
      first[prime[j] * i] = j;
      if (i % prime[j] == 0) break;
    }
  }
  pi[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pi[i] = pi[i - 1] * (long long)a[i] % MOD;
  }
  for (int i = 1; i < N; i++) rightmost[i] = N - 10;
  int num_q;
  cin >> num_q;
  num = 0;
  for (int i = 1; i <= num_q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    f[++num] = ones(l, r, i);
  }
  sort(f + 1, f + num + 1, cmp);
  for (int i = 0; i < N; i++) tree[i] = 1;
  int zz = n;
  for (int i = 1; i <= num; i++) {
    while (zz >= f[i].l) {
      add(zz);
      zz--;
    }
    long long tmp = pi[f[i].r] * mi(pi[f[i].l - 1], MOD - 2) % MOD;
    tmp = tmp * get(f[i].r) % MOD;
    ans[f[i].id] = tmp;
  }
  for (int i = 1; i <= num_q; i++) printf("%d\n", ans[i]);
  return 0;
}

```