#include <bits/stdc++.h>
using namespace std;
long long n, q, arr[200005], a, b, mod = 1e9 + 7, ans[200005], curl;
vector<long long> pos[1000005], rpos[200005];
unordered_set<long long> primes;
vector<pair<pair<long long, long long>, long long> > v;
long long qexp(long long a, long long b) {
  a %= mod;
  long long res = 1, base = a;
  while (b) {
    if (b & 1) {
      res *= base;
      res %= mod;
    }
    base *= base;
    base %= mod;
    b >>= 1;
  }
  return res % mod;
}
struct node {
  long long second, e, m, v;
  node *l, *r;
  node(long long S, long long E, long long st) {
    second = S, e = E, m = (second + e) / 2, v = 1;
    if (second == e) {
      if (second >= st)
        v = arr[second];
      else
        v = 1;
    } else {
      l = new node(second, m, st);
      r = new node(m + 1, e, st);
      v = (l->v * r->v) % mod;
    }
  }
  void up(long long x, long long nv) {
    if (second == e) {
      v = (v * nv) % mod;
      return;
    } else if (x <= m)
      l->up(x, nv);
    else
      r->up(x, nv);
    v = (l->v * r->v) % mod;
  }
  long long query(long long x, long long y) {
    if (second == x and e == y)
      return v;
    else if (y <= m)
      return l->query(x, y);
    else if (x > m)
      return r->query(x, y);
    else
      return (l->query(x, m) * r->query(m + 1, y)) % mod;
  }
} * seg;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = (long long)1; i <= (long long)n; i++) cin >> arr[i];
  cin >> q;
  for (long long i = (long long)1; i <= (long long)q; i++) {
    cin >> a >> b;
    v.push_back(pair<pair<long long, long long>, long long>(
        pair<long long, long long>(a, b), i));
  }
  sort(v.begin(), v.end());
  curl = v[0].first.first;
  seg = new node(1, n, curl);
  for (long long i = (long long)curl; i <= (long long)n; i++) {
    long long x = arr[i];
    for (long long j = (long long)2; j <= (long long)sqrt(x); j++) {
      if (x % j == 0) {
        pos[j].push_back(i);
        rpos[i].push_back(j);
        primes.insert(j);
      }
      while (x % j == 0) x /= j;
    }
    if (x >= 2) {
      pos[x].push_back(i);
      rpos[i].push_back(x);
      primes.insert(x);
    }
  }
  for (auto it : primes) {
    reverse(pos[it].begin(), pos[it].end());
    seg->up(pos[it].back(), ((it - 1) * qexp(it, mod - 2)) % mod);
  }
  for (auto it : v) {
    while (curl < it.first.first) {
      for (auto p : rpos[curl]) {
        seg->up(curl, (p * qexp(p - 1, mod - 2)) % mod);
        pos[p].pop_back();
        if (!pos[p].empty())
          seg->up(pos[p].back(), ((p - 1) * qexp(p, mod - 2)) % mod);
      }
      curl++;
    }
    ans[it.second] = seg->query(it.first.first, it.first.second);
  }
  for (long long i = (long long)1; i <= (long long)q; i++)
    cout << ans[i] << "\n";
}

### 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;
long long n, q, arr[200005], a, b, mod = 1e9 + 7, ans[200005], curl;
vector<long long> pos[1000005], rpos[200005];
unordered_set<long long> primes;
vector<pair<pair<long long, long long>, long long> > v;
long long qexp(long long a, long long b) {
  a %= mod;
  long long res = 1, base = a;
  while (b) {
    if (b & 1) {
      res *= base;
      res %= mod;
    }
    base *= base;
    base %= mod;
    b >>= 1;
  }
  return res % mod;
}
struct node {
  long long second, e, m, v;
  node *l, *r;
  node(long long S, long long E, long long st) {
    second = S, e = E, m = (second + e) / 2, v = 1;
    if (second == e) {
      if (second >= st)
        v = arr[second];
      else
        v = 1;
    } else {
      l = new node(second, m, st);
      r = new node(m + 1, e, st);
      v = (l->v * r->v) % mod;
    }
  }
  void up(long long x, long long nv) {
    if (second == e) {
      v = (v * nv) % mod;
      return;
    } else if (x <= m)
      l->up(x, nv);
    else
      r->up(x, nv);
    v = (l->v * r->v) % mod;
  }
  long long query(long long x, long long y) {
    if (second == x and e == y)
      return v;
    else if (y <= m)
      return l->query(x, y);
    else if (x > m)
      return r->query(x, y);
    else
      return (l->query(x, m) * r->query(m + 1, y)) % mod;
  }
} * seg;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = (long long)1; i <= (long long)n; i++) cin >> arr[i];
  cin >> q;
  for (long long i = (long long)1; i <= (long long)q; i++) {
    cin >> a >> b;
    v.push_back(pair<pair<long long, long long>, long long>(
        pair<long long, long long>(a, b), i));
  }
  sort(v.begin(), v.end());
  curl = v[0].first.first;
  seg = new node(1, n, curl);
  for (long long i = (long long)curl; i <= (long long)n; i++) {
    long long x = arr[i];
    for (long long j = (long long)2; j <= (long long)sqrt(x); j++) {
      if (x % j == 0) {
        pos[j].push_back(i);
        rpos[i].push_back(j);
        primes.insert(j);
      }
      while (x % j == 0) x /= j;
    }
    if (x >= 2) {
      pos[x].push_back(i);
      rpos[i].push_back(x);
      primes.insert(x);
    }
  }
  for (auto it : primes) {
    reverse(pos[it].begin(), pos[it].end());
    seg->up(pos[it].back(), ((it - 1) * qexp(it, mod - 2)) % mod);
  }
  for (auto it : v) {
    while (curl < it.first.first) {
      for (auto p : rpos[curl]) {
        seg->up(curl, (p * qexp(p - 1, mod - 2)) % mod);
        pos[p].pop_back();
        if (!pos[p].empty())
          seg->up(pos[p].back(), ((p - 1) * qexp(p, mod - 2)) % mod);
      }
      curl++;
    }
    ans[it.second] = seg->query(it.first.first, it.first.second);
  }
  for (long long i = (long long)1; i <= (long long)q; i++)
    cout << ans[i] << "\n";
}

```