#include <bits/stdc++.h>
using namespace std;
const int MaxN = 200010, MaxB = 500, MaxA = 1000010, mod = 1000000007;
struct Query {
  int l, r, id;
  bool operator<(const Query &a) const { return r < a.r; }
} q[MaxN];
int n, m, a[MaxN];
vector<pair<int, int> > f[MaxN];
int last[MaxA];
int prime[MaxA], ptot, minp[MaxA];
int inv[MaxA];
bool notp[MaxA];
void init(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!notp[i]) {
      prime[++ptot] = i;
      minp[i] = i;
    }
    for (int j = 1; j <= ptot && prime[j] * i <= n; ++j) {
      notp[prime[j] * i] = 1;
      minp[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  inv[1] = 1;
  for (int i = 2; i <= n; ++i)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
}
int ans[MaxN];
int c[MaxN];
void modify(int x, int v) {
  for (; x <= n; x += (x & -x)) c[x] = 1ll * c[x] * v % mod;
}
int query(int x) {
  int ret = 1;
  for (; x; x -= (x & -x)) ret = 1ll * ret * c[x] % mod;
  return ret;
}
int main() {
  scanf("%d", &n);
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    mx = max(mx, a[i]);
  }
  init(mx);
  for (int i = 1; i <= n; ++i) {
    int x = a[i];
    while (x != 1) {
      int prod = 1, t = minp[x];
      while (x % t == 0) {
        x /= t;
        prod *= t;
      }
      f[i].push_back(make_pair(t, prod));
    }
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + m + 1);
  for (int i = 0; i <= n; ++i) c[i] = 1;
  int node = 0;
  for (int i = 1; i <= n; ++i) {
    for (auto t : f[i]) {
      modify(1, t.second);
      modify(last[t.first] + 1, 1ll * (t.first - 1) * inv[t.first] % mod);
      modify(i + 1, inv[t.second / t.first * (t.first - 1)]);
      last[t.first] = i;
    }
    while (node < m && q[node + 1].r == i) {
      ++node;
      ans[q[node].id] = query(q[node].l);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please formulate 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 MaxN = 200010, MaxB = 500, MaxA = 1000010, mod = 1000000007;
struct Query {
  int l, r, id;
  bool operator<(const Query &a) const { return r < a.r; }
} q[MaxN];
int n, m, a[MaxN];
vector<pair<int, int> > f[MaxN];
int last[MaxA];
int prime[MaxA], ptot, minp[MaxA];
int inv[MaxA];
bool notp[MaxA];
void init(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!notp[i]) {
      prime[++ptot] = i;
      minp[i] = i;
    }
    for (int j = 1; j <= ptot && prime[j] * i <= n; ++j) {
      notp[prime[j] * i] = 1;
      minp[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  inv[1] = 1;
  for (int i = 2; i <= n; ++i)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
}
int ans[MaxN];
int c[MaxN];
void modify(int x, int v) {
  for (; x <= n; x += (x & -x)) c[x] = 1ll * c[x] * v % mod;
}
int query(int x) {
  int ret = 1;
  for (; x; x -= (x & -x)) ret = 1ll * ret * c[x] % mod;
  return ret;
}
int main() {
  scanf("%d", &n);
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    mx = max(mx, a[i]);
  }
  init(mx);
  for (int i = 1; i <= n; ++i) {
    int x = a[i];
    while (x != 1) {
      int prod = 1, t = minp[x];
      while (x % t == 0) {
        x /= t;
        prod *= t;
      }
      f[i].push_back(make_pair(t, prod));
    }
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + m + 1);
  for (int i = 0; i <= n; ++i) c[i] = 1;
  int node = 0;
  for (int i = 1; i <= n; ++i) {
    for (auto t : f[i]) {
      modify(1, t.second);
      modify(last[t.first] + 1, 1ll * (t.first - 1) * inv[t.first] % mod);
      modify(i + 1, inv[t.second / t.first * (t.first - 1)]);
      last[t.first] = i;
    }
    while (node < m && q[node + 1].r == i) {
      ++node;
      ans[q[node].id] = query(q[node].l);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```