#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long seg[4 * N];
long long a[N];
long long last[5 * N];
long long cur[5 * N];
vector<pair<long long, long long> > vec[N];
long long ans[N];
long long n, q;
void pre() {
  for (long long i = 0; i < 4 * N; i++) {
    seg[i] = 1;
  }
  for (long long i = 2; i < 5 * N; i++) {
    if (last[i] == 0) {
      for (long long j = i; j < 5 * N; j += i) {
        last[j] = i;
      }
    }
  }
}
void add(long long p, long long val, long long v = 1, long long s = 0,
         long long e = n) {
  seg[v] = (1LL) * seg[v] * val % mod;
  if (e - s == 1) {
    return;
  }
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  if (p < mid) {
    add(p, val, lc, s, mid);
  } else {
    add(p, val, rc, mid, e);
  }
}
long long get(long long l, long long r, long long v = 1, long long s = 0,
              long long e = n) {
  if (r <= s || e <= l) {
    return 1;
  }
  if (l <= s && e <= r) {
    return seg[v];
  }
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  return (get(l, r, lc, s, mid) * get(l, r, rc, mid, e)) % mod;
}
long long power(long long a, long long b) {
  if (!b) {
    return 1;
  }
  long long ans = power(a, b / 2);
  ans = ans * ans;
  ans = ans % mod;
  if (b % 2) {
    ans = ans * a;
    ans = ans % mod;
  }
  return ans;
}
int main() {
  pre();
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    vec[r].push_back({l, i});
  }
  memset(cur, -1, sizeof cur);
  for (long long i = 0; i < n; i++) {
    long long tmp = a[i];
    while (tmp > 1) {
      long long prv = last[tmp];
      long long cnt = 0;
      while (tmp % prv == 0) {
        tmp = tmp / prv;
        cnt++;
      }
      if (cur[prv] > -1) {
        long long tmp = prv * power(prv - 1, mod - 2) % mod;
        add(cur[prv], tmp);
      }
      long long tmp = (prv - 1) * power(prv, cnt - 1) % mod;
      add(i, tmp);
      cur[prv] = i;
    }
    for (long long j = 0; j < vec[i].size(); j++) {
      ans[vec[i][j].second] = get(vec[i][j].first, i + 1);
    }
  }
  for (long long i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
}

### Prompt

Generate 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 long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long seg[4 * N];
long long a[N];
long long last[5 * N];
long long cur[5 * N];
vector<pair<long long, long long> > vec[N];
long long ans[N];
long long n, q;
void pre() {
  for (long long i = 0; i < 4 * N; i++) {
    seg[i] = 1;
  }
  for (long long i = 2; i < 5 * N; i++) {
    if (last[i] == 0) {
      for (long long j = i; j < 5 * N; j += i) {
        last[j] = i;
      }
    }
  }
}
void add(long long p, long long val, long long v = 1, long long s = 0,
         long long e = n) {
  seg[v] = (1LL) * seg[v] * val % mod;
  if (e - s == 1) {
    return;
  }
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  if (p < mid) {
    add(p, val, lc, s, mid);
  } else {
    add(p, val, rc, mid, e);
  }
}
long long get(long long l, long long r, long long v = 1, long long s = 0,
              long long e = n) {
  if (r <= s || e <= l) {
    return 1;
  }
  if (l <= s && e <= r) {
    return seg[v];
  }
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  return (get(l, r, lc, s, mid) * get(l, r, rc, mid, e)) % mod;
}
long long power(long long a, long long b) {
  if (!b) {
    return 1;
  }
  long long ans = power(a, b / 2);
  ans = ans * ans;
  ans = ans % mod;
  if (b % 2) {
    ans = ans * a;
    ans = ans % mod;
  }
  return ans;
}
int main() {
  pre();
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    vec[r].push_back({l, i});
  }
  memset(cur, -1, sizeof cur);
  for (long long i = 0; i < n; i++) {
    long long tmp = a[i];
    while (tmp > 1) {
      long long prv = last[tmp];
      long long cnt = 0;
      while (tmp % prv == 0) {
        tmp = tmp / prv;
        cnt++;
      }
      if (cur[prv] > -1) {
        long long tmp = prv * power(prv - 1, mod - 2) % mod;
        add(cur[prv], tmp);
      }
      long long tmp = (prv - 1) * power(prv, cnt - 1) % mod;
      add(i, tmp);
      cur[prv] = i;
    }
    for (long long j = 0; j < vec[i].size(); j++) {
      ans[vec[i][j].second] = get(vec[i][j].first, i + 1);
    }
  }
  for (long long i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
}

```