#include <bits/stdc++.h>
using namespace std;
int n, sie[1000005], primes[1000005], prc, q, ans[200005], pind[200005];
vector<int> ot[80005], whp[200005];
int mulfen[200005], d[200005];
map<int, int> mp;
vector<pair<pair<int, int>, int> > vec;
int po(int x, int y) {
  if (y < 1) return 1;
  int ans = po(x, y / 2);
  ans = (ans * 1ll * ans) % 1000000007;
  if (y % 2) ans = (ans * 1ll * x) % 1000000007;
  return ans;
}
int rev(int x) { return po(x, 1000000007 - 2); }
void add(int pos, int val) {
  for (; pos <= n; pos += pos & -pos)
    mulfen[pos] = (mulfen[pos] * 1ll * val) % 1000000007;
}
int ask(int pos) {
  int val = 1;
  for (; pos > 0; pos -= pos & -pos)
    val = (val * 1ll * mulfen[pos]) % 1000000007;
  return val;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= 1000000; i++)
    if (sie[i] == 0) {
      primes[++prc] = i;
      mp[i] = prc;
      if (i < 1000)
        for (int j = i * i; j <= 1000000; j += i) {
          sie[j] = 1;
        }
    }
  for (int i = 1; i <= n; i++) mulfen[i] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d[i]);
    add(i, d[i]);
    int tem = d[i];
    for (int j = 1; primes[j] <= 1005; j++)
      if (tem % primes[j] == 0) {
        whp[i].push_back(j);
        ot[j].push_back(i);
        while (tem % primes[j] == 0) tem /= primes[j];
      }
    if (tem != 1) {
      ot[mp[tem]].push_back(i);
      whp[i].push_back(mp[tem]);
    }
  }
  for (int i = 1; i <= prc; i++)
    if (ot[i].size()) {
      add(ot[i][0], ((primes[i] - 1) * 1ll * rev(primes[i])) % 1000000007);
    }
  scanf("%d", &q);
  for (int i = 1, a, b; i <= q; i++) {
    scanf("%d %d", &a, &b);
    vec.push_back(pair<pair<int, int>, int>(pair<int, int>(a, b), i));
  }
  sort(vec.begin(), vec.end());
  for (int i = 1, vi = 0; i <= n && vi < (int)vec.size(); i++) {
    while (vi < (int)vec.size() && vec[vi].first.first == i) {
      ans[vec[vi].second] = ask(vec[vi].first.second);
      vi++;
    }
    add(i, rev(d[i]));
    for (int j = 0; j < (int)whp[i].size(); j++) {
      int curpind = whp[i][j];
      int div =
          ((primes[curpind] - 1) * 1ll * rev(primes[curpind])) % 1000000007;
      add(i, rev(div));
      pind[curpind]++;
      if (pind[curpind] < (int)ot[curpind].size())
        add(ot[curpind][pind[curpind]], div);
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}

### 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, sie[1000005], primes[1000005], prc, q, ans[200005], pind[200005];
vector<int> ot[80005], whp[200005];
int mulfen[200005], d[200005];
map<int, int> mp;
vector<pair<pair<int, int>, int> > vec;
int po(int x, int y) {
  if (y < 1) return 1;
  int ans = po(x, y / 2);
  ans = (ans * 1ll * ans) % 1000000007;
  if (y % 2) ans = (ans * 1ll * x) % 1000000007;
  return ans;
}
int rev(int x) { return po(x, 1000000007 - 2); }
void add(int pos, int val) {
  for (; pos <= n; pos += pos & -pos)
    mulfen[pos] = (mulfen[pos] * 1ll * val) % 1000000007;
}
int ask(int pos) {
  int val = 1;
  for (; pos > 0; pos -= pos & -pos)
    val = (val * 1ll * mulfen[pos]) % 1000000007;
  return val;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= 1000000; i++)
    if (sie[i] == 0) {
      primes[++prc] = i;
      mp[i] = prc;
      if (i < 1000)
        for (int j = i * i; j <= 1000000; j += i) {
          sie[j] = 1;
        }
    }
  for (int i = 1; i <= n; i++) mulfen[i] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d[i]);
    add(i, d[i]);
    int tem = d[i];
    for (int j = 1; primes[j] <= 1005; j++)
      if (tem % primes[j] == 0) {
        whp[i].push_back(j);
        ot[j].push_back(i);
        while (tem % primes[j] == 0) tem /= primes[j];
      }
    if (tem != 1) {
      ot[mp[tem]].push_back(i);
      whp[i].push_back(mp[tem]);
    }
  }
  for (int i = 1; i <= prc; i++)
    if (ot[i].size()) {
      add(ot[i][0], ((primes[i] - 1) * 1ll * rev(primes[i])) % 1000000007);
    }
  scanf("%d", &q);
  for (int i = 1, a, b; i <= q; i++) {
    scanf("%d %d", &a, &b);
    vec.push_back(pair<pair<int, int>, int>(pair<int, int>(a, b), i));
  }
  sort(vec.begin(), vec.end());
  for (int i = 1, vi = 0; i <= n && vi < (int)vec.size(); i++) {
    while (vi < (int)vec.size() && vec[vi].first.first == i) {
      ans[vec[vi].second] = ask(vec[vi].first.second);
      vi++;
    }
    add(i, rev(d[i]));
    for (int j = 0; j < (int)whp[i].size(); j++) {
      int curpind = whp[i][j];
      int div =
          ((primes[curpind] - 1) * 1ll * rev(primes[curpind])) % 1000000007;
      add(i, rev(div));
      pind[curpind]++;
      if (pind[curpind] < (int)ot[curpind].size())
        add(ot[curpind][pind[curpind]], div);
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}

```