#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7, N = 1e6 + 47, MAXN = 2e5 + 47;
long long prime[N], inv[N], fenw[MAXN], nxt[N];
void precalc() {
  for (long long i = 0; i < N; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (long long i = 2; i < N; i++) {
    if (prime[i] == 1) {
      for (long long j = i; j < N; j += i) prime[j] = i;
    }
  }
  inv[1] = 1;
  for (long long i = 2; i < N; ++i) inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;
  for (long long i = 0; i < MAXN; i++) fenw[i] = 1;
  for (long long i = 0; i < N; i++) nxt[i] = -1;
}
long long n;
void update(long long &ind, long long &add) {
  for (long long i = ind; i < n; i = ((i + 1) | i))
    fenw[i] = (fenw[i] * add) % MOD;
}
long long get_pr(long long r) {
  long long res = 1;
  for (; r >= 0; r = (r & (r + 1)) - 1) res = (res * fenw[r]) % MOD;
  return res;
}
vector<long long> a;
vector<vector<pair<long long, long long> > > quer;
void read() {
  precalc();
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void solve() {
  long long q;
  cin >> q;
  quer.resize(n);
  for (long long i = 0; i < q; i++) {
    long long a, b;
    cin >> a >> b;
    --a, --b;
    quer[a].push_back({b, i});
  }
  vector<long long> ans(q, 0);
  for (long long i = n - 1; i >= 0; i--) {
    update(i, a[i]);
    while (prime[a[i]] != 0) {
      long long dv = prime[a[i]];
      if (nxt[dv] != -1) {
        long long temp = (inv[dv - 1] * dv) % MOD;
        update(nxt[dv], temp);
      }
      nxt[dv] = i;
      long long temp = (inv[dv] * (dv - 1)) % MOD;
      update(i, temp);
      while (a[i] % dv == 0) a[i] /= dv;
    }
    for (pair<long long, long long> p : quer[i]) {
      ans[p.second] = get_pr(p.first);
    }
  }
  for (long long p : ans) cout << p << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    read();
    solve();
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MOD = 1e9 + 7, N = 1e6 + 47, MAXN = 2e5 + 47;
long long prime[N], inv[N], fenw[MAXN], nxt[N];
void precalc() {
  for (long long i = 0; i < N; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (long long i = 2; i < N; i++) {
    if (prime[i] == 1) {
      for (long long j = i; j < N; j += i) prime[j] = i;
    }
  }
  inv[1] = 1;
  for (long long i = 2; i < N; ++i) inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;
  for (long long i = 0; i < MAXN; i++) fenw[i] = 1;
  for (long long i = 0; i < N; i++) nxt[i] = -1;
}
long long n;
void update(long long &ind, long long &add) {
  for (long long i = ind; i < n; i = ((i + 1) | i))
    fenw[i] = (fenw[i] * add) % MOD;
}
long long get_pr(long long r) {
  long long res = 1;
  for (; r >= 0; r = (r & (r + 1)) - 1) res = (res * fenw[r]) % MOD;
  return res;
}
vector<long long> a;
vector<vector<pair<long long, long long> > > quer;
void read() {
  precalc();
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void solve() {
  long long q;
  cin >> q;
  quer.resize(n);
  for (long long i = 0; i < q; i++) {
    long long a, b;
    cin >> a >> b;
    --a, --b;
    quer[a].push_back({b, i});
  }
  vector<long long> ans(q, 0);
  for (long long i = n - 1; i >= 0; i--) {
    update(i, a[i]);
    while (prime[a[i]] != 0) {
      long long dv = prime[a[i]];
      if (nxt[dv] != -1) {
        long long temp = (inv[dv - 1] * dv) % MOD;
        update(nxt[dv], temp);
      }
      nxt[dv] = i;
      long long temp = (inv[dv] * (dv - 1)) % MOD;
      update(i, temp);
      while (a[i] % dv == 0) a[i] /= dv;
    }
    for (pair<long long, long long> p : quer[i]) {
      ans[p.second] = get_pr(p.first);
    }
  }
  for (long long p : ans) cout << p << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    read();
    solve();
  }
}

```