#include <bits/stdc++.h>
const double PI = acos(-1.0);
const int DIR = 4;
const int X[] = {1, 0, -1, 0};
const int Y[] = {0, 1, 0, -1};
const int N = (int)2e5 + 228;
const int MOD = (int)1e9 + 7;
const int M = (int)1e6 + 228;
using namespace std;
int n, Q, pref[M], lp[M], L[N], a[N];
long long t[N * 4], ans[N], f[N], mu[M], di[M];
vector<pair<int, int> > q[N];
vector<int> p[M];
long long binpow(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * a % MOD;
    a = a * a % MOD;
    k >>= 1;
  }
  return res;
}
void build() {
  for (int v = 1; v <= n + n - 1; v++) t[v] = 1;
}
void upd(int v, long long val) {
  v += n - 1;
  t[v] = t[v] * val % MOD;
  while (v != 1) {
    v >>= 1;
    t[v] = t[v + v] * t[v + v + 1] % MOD;
  }
}
long long get(int l, int r) {
  l += n - 1, r += n - 1;
  long long res = 1;
  while (l <= r) {
    if (l & 1) res = res * t[l++] % MOD;
    if (!(r & 1)) res = res * t[r--] % MOD;
    l >>= 1, r >>= 1;
  }
  return res;
}
int main() {
  for (int i = 2; i < M; i++) {
    if (lp[i] == 0) {
      mu[i] = (1 - binpow(i, MOD - 2) + MOD) % MOD;
      di[i] = binpow(mu[i], MOD - 2);
      lp[i] = i;
      if (1ll * i * i < M)
        for (int j = i * i; j < M; j += i) lp[j] = i;
    }
  }
  f[0] = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * a[i] % MOD;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    L[i] = l;
    q[r].push_back({1, i});
    ans[i] = f[r] * binpow(f[l - 1], MOD - 2) % MOD;
  }
  build();
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    while (x != 1) {
      int y = lp[x];
      while (x % y == 0) x /= y;
      upd(i, mu[y]);
      if (pref[y]) upd(pref[y], di[y]);
      pref[y] = i;
    }
    for (auto it : q[i]) {
      if (it.first)
        ans[it.second] = ans[it.second] * get(L[it.second], i) % MOD;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%lld\n", ans[i]);
  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>
const double PI = acos(-1.0);
const int DIR = 4;
const int X[] = {1, 0, -1, 0};
const int Y[] = {0, 1, 0, -1};
const int N = (int)2e5 + 228;
const int MOD = (int)1e9 + 7;
const int M = (int)1e6 + 228;
using namespace std;
int n, Q, pref[M], lp[M], L[N], a[N];
long long t[N * 4], ans[N], f[N], mu[M], di[M];
vector<pair<int, int> > q[N];
vector<int> p[M];
long long binpow(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * a % MOD;
    a = a * a % MOD;
    k >>= 1;
  }
  return res;
}
void build() {
  for (int v = 1; v <= n + n - 1; v++) t[v] = 1;
}
void upd(int v, long long val) {
  v += n - 1;
  t[v] = t[v] * val % MOD;
  while (v != 1) {
    v >>= 1;
    t[v] = t[v + v] * t[v + v + 1] % MOD;
  }
}
long long get(int l, int r) {
  l += n - 1, r += n - 1;
  long long res = 1;
  while (l <= r) {
    if (l & 1) res = res * t[l++] % MOD;
    if (!(r & 1)) res = res * t[r--] % MOD;
    l >>= 1, r >>= 1;
  }
  return res;
}
int main() {
  for (int i = 2; i < M; i++) {
    if (lp[i] == 0) {
      mu[i] = (1 - binpow(i, MOD - 2) + MOD) % MOD;
      di[i] = binpow(mu[i], MOD - 2);
      lp[i] = i;
      if (1ll * i * i < M)
        for (int j = i * i; j < M; j += i) lp[j] = i;
    }
  }
  f[0] = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * a[i] % MOD;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    L[i] = l;
    q[r].push_back({1, i});
    ans[i] = f[r] * binpow(f[l - 1], MOD - 2) % MOD;
  }
  build();
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    while (x != 1) {
      int y = lp[x];
      while (x % y == 0) x /= y;
      upd(i, mu[y]);
      if (pref[y]) upd(pref[y], di[y]);
      pref[y] = i;
    }
    for (auto it : q[i]) {
      if (it.first)
        ans[it.second] = ans[it.second] * get(L[it.second], i) % MOD;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%lld\n", ans[i]);
  return 0;
}

```