#include <bits/stdc++.h>
using namespace std;
int n, q, tot, vis[200010 * 5];
long long Max, p[200010], a[200010], val[200010], inv[200010], ans[200010],
    tr[200010 * 4];
vector<int> fac[200010 * 5];
vector<pair<int, int> > que[200010];
void build(int p, int L, int R) {
  if (L == R) {
    tr[p] = a[L];
    return;
  }
  build(p << 1, L, ((L + R) >> 1));
  build(p << 1 | 1, ((L + R) >> 1) + 1, R);
  tr[p] = tr[p << 1] * tr[p << 1 | 1] % 1000000007;
}
void modify(int p, int L, int R, int pos, long long w) {
  if (L == R) {
    tr[p] = tr[p] * w % 1000000007;
    return;
  }
  if (pos <= ((L + R) >> 1))
    modify(p << 1, L, ((L + R) >> 1), pos, w);
  else
    modify(p << 1 | 1, ((L + R) >> 1) + 1, R, pos, w);
  tr[p] = tr[p << 1] * tr[p << 1 | 1] % 1000000007;
}
long long query(int p, int L, int R, int l, int r) {
  if (l <= L && R <= r) return tr[p];
  long long res = 1;
  if (l <= ((L + R) >> 1))
    res = res * query(p << 1, L, ((L + R) >> 1), l, r) % 1000000007;
  if (r > ((L + R) >> 1))
    res = res * query(p << 1 | 1, ((L + R) >> 1) + 1, R, l, r) % 1000000007;
  return res;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), Max = max(Max, a[i]);
  for (int i = 2; i <= Max; i++)
    if (!vis[i]) {
      p[++tot] = i;
      for (int j = 1; i * j <= Max; j++)
        vis[i * j] = 1, fac[i * j].push_back(tot);
    }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= tot; i++) {
    val[i] = (p[i] - 1) * power(p[i], 1000000007 - 2) % 1000000007;
    inv[i] = power(val[i], 1000000007 - 2);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    que[l].push_back(make_pair(i, r));
  }
  build(1, 1, n);
  for (int i = n; i >= 1; i--) {
    for (int j : fac[a[i]]) {
      if (vis[j]) modify(1, 1, n, vis[j], inv[j]);
      vis[j] = i;
      modify(1, 1, n, i, val[j]);
    }
    for (auto j : que[i]) ans[j.first] = query(1, 1, n, i, j.second);
  }
  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>
using namespace std;
int n, q, tot, vis[200010 * 5];
long long Max, p[200010], a[200010], val[200010], inv[200010], ans[200010],
    tr[200010 * 4];
vector<int> fac[200010 * 5];
vector<pair<int, int> > que[200010];
void build(int p, int L, int R) {
  if (L == R) {
    tr[p] = a[L];
    return;
  }
  build(p << 1, L, ((L + R) >> 1));
  build(p << 1 | 1, ((L + R) >> 1) + 1, R);
  tr[p] = tr[p << 1] * tr[p << 1 | 1] % 1000000007;
}
void modify(int p, int L, int R, int pos, long long w) {
  if (L == R) {
    tr[p] = tr[p] * w % 1000000007;
    return;
  }
  if (pos <= ((L + R) >> 1))
    modify(p << 1, L, ((L + R) >> 1), pos, w);
  else
    modify(p << 1 | 1, ((L + R) >> 1) + 1, R, pos, w);
  tr[p] = tr[p << 1] * tr[p << 1 | 1] % 1000000007;
}
long long query(int p, int L, int R, int l, int r) {
  if (l <= L && R <= r) return tr[p];
  long long res = 1;
  if (l <= ((L + R) >> 1))
    res = res * query(p << 1, L, ((L + R) >> 1), l, r) % 1000000007;
  if (r > ((L + R) >> 1))
    res = res * query(p << 1 | 1, ((L + R) >> 1) + 1, R, l, r) % 1000000007;
  return res;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), Max = max(Max, a[i]);
  for (int i = 2; i <= Max; i++)
    if (!vis[i]) {
      p[++tot] = i;
      for (int j = 1; i * j <= Max; j++)
        vis[i * j] = 1, fac[i * j].push_back(tot);
    }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= tot; i++) {
    val[i] = (p[i] - 1) * power(p[i], 1000000007 - 2) % 1000000007;
    inv[i] = power(val[i], 1000000007 - 2);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    que[l].push_back(make_pair(i, r));
  }
  build(1, 1, n);
  for (int i = n; i >= 1; i--) {
    for (int j : fac[a[i]]) {
      if (vis[j]) modify(1, 1, n, vis[j], inv[j]);
      vis[j] = i;
      modify(1, 1, n, i, val[j]);
    }
    for (auto j : que[i]) ans[j.first] = query(1, 1, n, i, j.second);
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}

```