#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
const int MOD = 1e9 + 7;
long long ksm(long long a, long long b, long long mod) {
  long long res = 1, base = a;
  while (b) {
    if (b & 1) res = (res * base) % mod;
    base = (base * base) % mod;
    b >>= 1;
  }
  return res;
}
vector<long long> prime;
bool vis[maxn];
vector<int> factor[maxn];
void PHI(int n) {
  memset(vis, 1, sizeof(vis));
  for (int i = 2; i <= n; i++) {
    if (vis[i]) prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = 0;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 0; i < prime.size(); i++) {
    for (int k = 1; k * prime[i] <= n; k++) {
      factor[k * prime[i]].push_back(prime[i]);
    }
  }
}
int N, Q;
long long a[maxn], sum[maxn] = {1};
struct AAA {
  int L, R, id;
  bool operator<(const AAA &tp) const { return R < tp.R; }
};
AAA q[maxn];
long long tree[maxn];
void Add(int i, long long val) {
  if (i < 1) return;
  for (; i < maxn; i += i & -i) tree[i] = (tree[i] * val) % MOD;
}
long long getsum(int i) {
  long long res = 1;
  for (; i >= 1; i -= i & -i) res = (res * tree[i]) % MOD;
  return res;
}
int main() {
  PHI(maxn - 5);
  while (cin >> N) {
    for (int i = 0; i < maxn; i++) tree[i] = 1;
    for (int i = 1; i <= N; i++) {
      scanf("%lld", a + i);
      sum[i] = sum[i - 1] * a[i] % MOD;
    }
    cin >> Q;
    for (int i = 1; i <= Q; i++) {
      scanf("%d%d", &q[i].L, &q[i].R);
      q[i].id = i;
    }
    sort(q + 1, q + 1 + Q);
    static int last[maxn];
    memset(last, 0, sizeof last);
    static int ans[maxn];
    int now = 1;
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j < factor[a[i]].size(); j++) {
        long long p = factor[a[i]][j];
        if (last[p]) Add(last[p], p * ksm(p - 1, MOD - 2, MOD) % MOD);
        Add(i, (long long)(p - 1) * ksm(p, MOD - 2, MOD) % MOD);
        last[p] = i;
      }
      while (q[now].R == i) {
        int id = q[now].id;
        int qL = q[now].L;
        ans[id] = sum[i] * ksm(sum[qL - 1], MOD - 2, MOD) % MOD * getsum(i) %
                  MOD * ksm(getsum(qL - 1), MOD - 2, MOD) % MOD;
        now++;
      }
    }
    for (int i = 1; i <= Q; i++) cout << ans[i] << endl;
  }
}

### 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;
const int maxn = 1e6 + 5;
const int MOD = 1e9 + 7;
long long ksm(long long a, long long b, long long mod) {
  long long res = 1, base = a;
  while (b) {
    if (b & 1) res = (res * base) % mod;
    base = (base * base) % mod;
    b >>= 1;
  }
  return res;
}
vector<long long> prime;
bool vis[maxn];
vector<int> factor[maxn];
void PHI(int n) {
  memset(vis, 1, sizeof(vis));
  for (int i = 2; i <= n; i++) {
    if (vis[i]) prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = 0;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 0; i < prime.size(); i++) {
    for (int k = 1; k * prime[i] <= n; k++) {
      factor[k * prime[i]].push_back(prime[i]);
    }
  }
}
int N, Q;
long long a[maxn], sum[maxn] = {1};
struct AAA {
  int L, R, id;
  bool operator<(const AAA &tp) const { return R < tp.R; }
};
AAA q[maxn];
long long tree[maxn];
void Add(int i, long long val) {
  if (i < 1) return;
  for (; i < maxn; i += i & -i) tree[i] = (tree[i] * val) % MOD;
}
long long getsum(int i) {
  long long res = 1;
  for (; i >= 1; i -= i & -i) res = (res * tree[i]) % MOD;
  return res;
}
int main() {
  PHI(maxn - 5);
  while (cin >> N) {
    for (int i = 0; i < maxn; i++) tree[i] = 1;
    for (int i = 1; i <= N; i++) {
      scanf("%lld", a + i);
      sum[i] = sum[i - 1] * a[i] % MOD;
    }
    cin >> Q;
    for (int i = 1; i <= Q; i++) {
      scanf("%d%d", &q[i].L, &q[i].R);
      q[i].id = i;
    }
    sort(q + 1, q + 1 + Q);
    static int last[maxn];
    memset(last, 0, sizeof last);
    static int ans[maxn];
    int now = 1;
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j < factor[a[i]].size(); j++) {
        long long p = factor[a[i]][j];
        if (last[p]) Add(last[p], p * ksm(p - 1, MOD - 2, MOD) % MOD);
        Add(i, (long long)(p - 1) * ksm(p, MOD - 2, MOD) % MOD);
        last[p] = i;
      }
      while (q[now].R == i) {
        int id = q[now].id;
        int qL = q[now].L;
        ans[id] = sum[i] * ksm(sum[qL - 1], MOD - 2, MOD) % MOD * getsum(i) %
                  MOD * ksm(getsum(qL - 1), MOD - 2, MOD) % MOD;
        now++;
      }
    }
    for (int i = 1; i <= Q; i++) cout << ans[i] << endl;
  }
}

```