#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200002, MN = 1000002;
const long long mod = 1000000007;
vector<pair<int, int> > Q[MAXN];
int a[MAXN], ans[MAXN];
int pos[MN], p[MN], n, q;
long long tree[MAXN * 2];
void upd(int l, int r, int x) {
  for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) tree[l] = tree[l] * x % mod, l++;
    if (r & 1) --r, tree[r] = tree[r] * x % mod;
  }
}
long long query(int i) {
  int ret = 1;
  for (i += n; i > 0; i >>= 1) ret = ret * tree[i] % mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(pos, -1, sizeof pos);
  for (int i = 2; i * i < MN; i++)
    if (!p[i])
      for (int j = i * i; j < MN; j += i) p[j] = i;
  for (int i = 1; i < MN; i++)
    if (!p[i]) p[i] = i;
  cin >> n;
  for (int i = 0; i < n + n; i++) tree[i] = 1;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1) {
    int r = a[0];
    while (a[0] > 1) {
      int t = p[a[0]];
      while (a[0] % t == 0) a[0] /= t;
      r /= t;
      r *= t - 1;
    }
    cin >> q;
    for (int i = 0; i < q; i++) cout << r << '\n';
    return 0;
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    Q[r - 1].push_back({l - 1, i});
  }
  for (int i = 0; i < n; i++) {
    int x = a[i];
    while (x > 1) {
      int r = p[x], m = 1;
      while (x % r == 0) x /= r, m *= r;
      if (m != r) upd(0, i, m / r);
      upd(pos[r] + 1, i, r - 1);
      upd(0, pos[r], r);
      pos[r] = i;
    }
    for (auto x : Q[i]) ans[x.second] = query(x.first);
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
}

### Prompt

Create a solution in cpp for 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 = 200002, MN = 1000002;
const long long mod = 1000000007;
vector<pair<int, int> > Q[MAXN];
int a[MAXN], ans[MAXN];
int pos[MN], p[MN], n, q;
long long tree[MAXN * 2];
void upd(int l, int r, int x) {
  for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) tree[l] = tree[l] * x % mod, l++;
    if (r & 1) --r, tree[r] = tree[r] * x % mod;
  }
}
long long query(int i) {
  int ret = 1;
  for (i += n; i > 0; i >>= 1) ret = ret * tree[i] % mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(pos, -1, sizeof pos);
  for (int i = 2; i * i < MN; i++)
    if (!p[i])
      for (int j = i * i; j < MN; j += i) p[j] = i;
  for (int i = 1; i < MN; i++)
    if (!p[i]) p[i] = i;
  cin >> n;
  for (int i = 0; i < n + n; i++) tree[i] = 1;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1) {
    int r = a[0];
    while (a[0] > 1) {
      int t = p[a[0]];
      while (a[0] % t == 0) a[0] /= t;
      r /= t;
      r *= t - 1;
    }
    cin >> q;
    for (int i = 0; i < q; i++) cout << r << '\n';
    return 0;
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    Q[r - 1].push_back({l - 1, i});
  }
  for (int i = 0; i < n; i++) {
    int x = a[i];
    while (x > 1) {
      int r = p[x], m = 1;
      while (x % r == 0) x /= r, m *= r;
      if (m != r) upd(0, i, m / r);
      upd(pos[r] + 1, i, r - 1);
      upd(0, pos[r], r);
      pos[r] = i;
    }
    for (auto x : Q[i]) ans[x.second] = query(x.first);
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
}

```