#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAXN = 200 * 1000 + 3, D = 1000 * 1000 * 1000 + 7,
                MAXX = 1000 * 1000 + 3, MXCP = 7 + 1, MAXS = MXCP * MAXN + 3;
inline long long inv(long long x) {
  long long ret = 1;
  for (long long i = D - 2; i; (i >>= 1), (x = x * x % D))
    if (i & 1) ret = ret * x % D;
  return ret;
}
long long n, q, fen[MAXS], nxt[MAXS], hlp[MAXX], r[MAXN], a[MAXN], ans[MAXN],
    b[MAXS], ptb, tl[MAXN], tr[MAXN], rr[MAXN], rl[MAXN];
vector<long long> qv[MAXS], pm;
inline void inpre() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    tl[i] = ptb;
    cin >> a[i];
    long long x = a[i];
    for (long long j = 2; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        b[ptb++] = j;
        while (a[i] % j == 0) a[i] /= j;
      }
    if (a[i] > 1) b[ptb++] = a[i];
    tr[i] = ptb;
    a[i] = x;
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> rl[i] >> rr[i];
    r[i] = tr[--rr[i]];
    rr[i]++;
    qv[tl[--rl[i]]].push_back(i);
  }
  memset(hlp, -1, sizeof hlp);
  for (long long i = ptb - 1; ~i; i--) {
    nxt[i] = hlp[b[i]];
    hlp[b[i]] = i;
  }
  pm.push_back(1);
  for (long long i = 0; i < n; i++) {
    long long tmp = pm.back() * a[i] % D;
    pm.push_back(tmp);
  }
}
inline void upd(long long p, long long val) {
  for (p++; p <= ptb; p += (p & (-p))) (fen[p] *= val) %= D;
}
inline long long get(long long p) {
  long long ret = 1;
  for (; p; p -= (p & (-p))) (ret *= fen[p]) %= D;
  return ret;
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  inpre();
  fill(fen, fen + MAXS, 1);
  for (long long i = 0; i < MAXX; i++)
    if (~hlp[i]) upd(hlp[i], (1 - inv(i) + D) % D);
  for (long long i = 0; i < ptb; i++) {
    for (auto j : qv[i])
      ans[j] = get(r[j]) * pm[rr[j]] % D * inv(pm[rl[j]]) % D;
    upd(i, inv((1 - inv(b[i]) + D) % D));
    if (~nxt[i]) upd(nxt[i], (1 - inv(b[i]) + D) % D);
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAXN = 200 * 1000 + 3, D = 1000 * 1000 * 1000 + 7,
                MAXX = 1000 * 1000 + 3, MXCP = 7 + 1, MAXS = MXCP * MAXN + 3;
inline long long inv(long long x) {
  long long ret = 1;
  for (long long i = D - 2; i; (i >>= 1), (x = x * x % D))
    if (i & 1) ret = ret * x % D;
  return ret;
}
long long n, q, fen[MAXS], nxt[MAXS], hlp[MAXX], r[MAXN], a[MAXN], ans[MAXN],
    b[MAXS], ptb, tl[MAXN], tr[MAXN], rr[MAXN], rl[MAXN];
vector<long long> qv[MAXS], pm;
inline void inpre() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    tl[i] = ptb;
    cin >> a[i];
    long long x = a[i];
    for (long long j = 2; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        b[ptb++] = j;
        while (a[i] % j == 0) a[i] /= j;
      }
    if (a[i] > 1) b[ptb++] = a[i];
    tr[i] = ptb;
    a[i] = x;
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> rl[i] >> rr[i];
    r[i] = tr[--rr[i]];
    rr[i]++;
    qv[tl[--rl[i]]].push_back(i);
  }
  memset(hlp, -1, sizeof hlp);
  for (long long i = ptb - 1; ~i; i--) {
    nxt[i] = hlp[b[i]];
    hlp[b[i]] = i;
  }
  pm.push_back(1);
  for (long long i = 0; i < n; i++) {
    long long tmp = pm.back() * a[i] % D;
    pm.push_back(tmp);
  }
}
inline void upd(long long p, long long val) {
  for (p++; p <= ptb; p += (p & (-p))) (fen[p] *= val) %= D;
}
inline long long get(long long p) {
  long long ret = 1;
  for (; p; p -= (p & (-p))) (ret *= fen[p]) %= D;
  return ret;
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  inpre();
  fill(fen, fen + MAXS, 1);
  for (long long i = 0; i < MAXX; i++)
    if (~hlp[i]) upd(hlp[i], (1 - inv(i) + D) % D);
  for (long long i = 0; i < ptb; i++) {
    for (auto j : qv[i])
      ans[j] = get(r[j]) * pm[rr[j]] % D * inv(pm[rl[j]]) % D;
    upd(i, inv((1 - inv(b[i]) + D) % D));
    if (~nxt[i]) upd(nxt[i], (1 - inv(b[i]) + D) % D);
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}

```