#include <bits/stdc++.h>
using namespace std;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 7;
long long n, a[N], q, fenwick[N], pos, ans[N];
pair<int, pair<int, int>> interval[N];
vector<int> p[N];
bool cmp(pair<int, pair<int, int>> lhs, pair<int, pair<int, int>> rhs) {
  return lhs.second.second < rhs.second.second;
}
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % MOD, b >> 1) * ((b & 1) ? a : 1)) % MOD : 1;
}
void update(long long pos, long long val) {
  for (int i = pos; i <= n; i += i & (-i))
    fenwick[i] = (fenwick[i] * val) % MOD;
}
void update2(long long pos, long long val) {
  long long number = mypow(val, MOD - 2);
  for (int i = pos; i <= n; i += i & (-i))
    fenwick[i] = (fenwick[i] * number) % MOD;
}
void upd(int pos, int num) {
  for (int i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      long long z = 1;
      while (num % i == 0) {
        num /= i;
        z *= i;
      }
      z /= i;
      if (p[i].size()) {
        update2(p[i][p[i].size() - 1], i - 1);
        update(p[i][p[i].size() - 1], i);
      }
      update(pos, (z * (i - 1)) % MOD);
      p[i].push_back(pos);
    }
  }
  if (num > 1) {
    if (p[num].size()) {
      update2(p[num][p[num].size() - 1], num - 1);
      update(p[num][p[num].size() - 1], num);
    }
    update(pos, num - 1);
    p[num].push_back(pos);
  }
}
long long mul(int e) {
  long long func_ans = 1;
  for (int i = e; i > 0; i -= i & (-i))
    func_ans = (func_ans * fenwick[i]) % MOD;
  return func_ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < N; i++) fenwick[i] = 1;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> interval[i].second.first >> interval[i].second.second;
    interval[i].first = i;
  }
  sort(interval, interval + q, cmp);
  for (int i = 1; i <= n; i++) {
    upd(i, a[i]);
    while (interval[pos].second.second == i) {
      ans[interval[pos].first] =
          (mul(i) * mypow(mul(interval[pos].second.first - 1), MOD - 2)) % MOD;
      pos++;
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  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;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 7;
long long n, a[N], q, fenwick[N], pos, ans[N];
pair<int, pair<int, int>> interval[N];
vector<int> p[N];
bool cmp(pair<int, pair<int, int>> lhs, pair<int, pair<int, int>> rhs) {
  return lhs.second.second < rhs.second.second;
}
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % MOD, b >> 1) * ((b & 1) ? a : 1)) % MOD : 1;
}
void update(long long pos, long long val) {
  for (int i = pos; i <= n; i += i & (-i))
    fenwick[i] = (fenwick[i] * val) % MOD;
}
void update2(long long pos, long long val) {
  long long number = mypow(val, MOD - 2);
  for (int i = pos; i <= n; i += i & (-i))
    fenwick[i] = (fenwick[i] * number) % MOD;
}
void upd(int pos, int num) {
  for (int i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      long long z = 1;
      while (num % i == 0) {
        num /= i;
        z *= i;
      }
      z /= i;
      if (p[i].size()) {
        update2(p[i][p[i].size() - 1], i - 1);
        update(p[i][p[i].size() - 1], i);
      }
      update(pos, (z * (i - 1)) % MOD);
      p[i].push_back(pos);
    }
  }
  if (num > 1) {
    if (p[num].size()) {
      update2(p[num][p[num].size() - 1], num - 1);
      update(p[num][p[num].size() - 1], num);
    }
    update(pos, num - 1);
    p[num].push_back(pos);
  }
}
long long mul(int e) {
  long long func_ans = 1;
  for (int i = e; i > 0; i -= i & (-i))
    func_ans = (func_ans * fenwick[i]) % MOD;
  return func_ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < N; i++) fenwick[i] = 1;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> interval[i].second.first >> interval[i].second.second;
    interval[i].first = i;
  }
  sort(interval, interval + q, cmp);
  for (int i = 1; i <= n; i++) {
    upd(i, a[i]);
    while (interval[pos].second.second == i) {
      ans[interval[pos].first] =
          (mul(i) * mypow(mul(interval[pos].second.first - 1), MOD - 2)) % MOD;
      pos++;
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}

```