#include <bits/stdc++.h>
using namespace std;
const long long MAX1 = 2e5 + 5;
const long long MAX2 = 1e6 + 5;
const long long MOD = 1e9 + 7;
long long n, q;
long long arr[MAX1];
vector<pair<pair<long long, long long>, long long>> que;
map<long long, deque<long long>> pos;
vector<pair<long long, long long>> ans;
vector<long long> fp[MAX1];
long long primo[MAX2];
long long tree[MAX1 * 4];
long long find_id[MAX1];
void criba() {
  for (long long i = 1; i < (long long)1e6 + 1; i++) primo[i] = i;
  for (long long i = 2; i < (long long)1e6 + 1; i++) {
    if (primo[i] == i && i <= 1e3)
      for (long long j = i * i; j <= 1e6; j += i) primo[j] = i;
  }
}
void fact(long long x, long long idx) {
  while (x > 1) {
    long long a = primo[x];
    pos[a].push_back(idx);
    fp[idx].push_back(a);
    while (x % a == 0) x /= a;
  }
}
long long op(long long a, long long b) { return (a * b) % MOD; }
void build(long long id, long long l, long long r) {
  if (r == l + 1) {
    tree[id] = arr[l];
    find_id[l] = id;
    return;
  }
  long long med = (l + r) >> 1;
  build(id << 1, l, med);
  build(id << 1 | 1, med, r);
  tree[id] = op(tree[id << 1], tree[id << 1 | 1]);
}
void up(long long nod, long long a, long long b) {
  long long id = find_id[nod];
  tree[id] /= a;
  tree[id] *= b;
  tree[id] = tree[id] % MOD;
  while (id > 0) {
    id >>= 1;
    tree[id] = op(tree[id << 1], tree[id << 1 | 1]);
  }
}
long long query(long long a, long long b, long long id, long long l,
                long long r) {
  if (r <= a or b <= l) return 1;
  if (a <= l and r <= b) {
    return tree[id];
  }
  long long med = (l + r) >> 1;
  long long izq = query(a, b, id << 1, l, med);
  long long der = query(a, b, id << 1 | 1, med, r);
  return op(izq, der);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) cin >> arr[i];
  cin >> q;
  criba();
  for (long long i = 0; i < (long long)n; i++) fact(arr[i], i);
  for (long long i = 0; i < (long long)q; i++) {
    long long a, b;
    long long idx = (long long)i;
    cin >> a >> b;
    a--;
    que.push_back({{a, b}, idx});
  }
  sort((que).begin(), (que).end());
  build(1, 0, n);
  for (auto it : pos) {
    long long a = it.second.front();
    up(a, it.first, it.first - 1);
  }
  long long f1 = 0;
  for (long long i = 0; i < (long long)que.size(); i++) {
    long long x = que[i].first.first;
    long long y = que[i].first.second;
    long long id = que[i].second;
    if (f1 < x) {
      for (long long j = f1; j < (long long)x; j++) {
        for (long long z = 0; z < (long long)fp[j].size(); z++) {
          int pri = fp[j][z];
          pos[pri].pop_front();
          if (!pos[pri].empty()) {
            up(pos[pri].front(), pri, pri - 1);
          }
        }
      }
      f1 = x;
    }
    ans.push_back({id, query(x, y, 1, 0, n)});
  }
  sort((ans).begin(), (ans).end());
  for (long long i = 0; i < (long long)ans.size(); i++) {
    cout << ans[i].second << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 MAX1 = 2e5 + 5;
const long long MAX2 = 1e6 + 5;
const long long MOD = 1e9 + 7;
long long n, q;
long long arr[MAX1];
vector<pair<pair<long long, long long>, long long>> que;
map<long long, deque<long long>> pos;
vector<pair<long long, long long>> ans;
vector<long long> fp[MAX1];
long long primo[MAX2];
long long tree[MAX1 * 4];
long long find_id[MAX1];
void criba() {
  for (long long i = 1; i < (long long)1e6 + 1; i++) primo[i] = i;
  for (long long i = 2; i < (long long)1e6 + 1; i++) {
    if (primo[i] == i && i <= 1e3)
      for (long long j = i * i; j <= 1e6; j += i) primo[j] = i;
  }
}
void fact(long long x, long long idx) {
  while (x > 1) {
    long long a = primo[x];
    pos[a].push_back(idx);
    fp[idx].push_back(a);
    while (x % a == 0) x /= a;
  }
}
long long op(long long a, long long b) { return (a * b) % MOD; }
void build(long long id, long long l, long long r) {
  if (r == l + 1) {
    tree[id] = arr[l];
    find_id[l] = id;
    return;
  }
  long long med = (l + r) >> 1;
  build(id << 1, l, med);
  build(id << 1 | 1, med, r);
  tree[id] = op(tree[id << 1], tree[id << 1 | 1]);
}
void up(long long nod, long long a, long long b) {
  long long id = find_id[nod];
  tree[id] /= a;
  tree[id] *= b;
  tree[id] = tree[id] % MOD;
  while (id > 0) {
    id >>= 1;
    tree[id] = op(tree[id << 1], tree[id << 1 | 1]);
  }
}
long long query(long long a, long long b, long long id, long long l,
                long long r) {
  if (r <= a or b <= l) return 1;
  if (a <= l and r <= b) {
    return tree[id];
  }
  long long med = (l + r) >> 1;
  long long izq = query(a, b, id << 1, l, med);
  long long der = query(a, b, id << 1 | 1, med, r);
  return op(izq, der);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) cin >> arr[i];
  cin >> q;
  criba();
  for (long long i = 0; i < (long long)n; i++) fact(arr[i], i);
  for (long long i = 0; i < (long long)q; i++) {
    long long a, b;
    long long idx = (long long)i;
    cin >> a >> b;
    a--;
    que.push_back({{a, b}, idx});
  }
  sort((que).begin(), (que).end());
  build(1, 0, n);
  for (auto it : pos) {
    long long a = it.second.front();
    up(a, it.first, it.first - 1);
  }
  long long f1 = 0;
  for (long long i = 0; i < (long long)que.size(); i++) {
    long long x = que[i].first.first;
    long long y = que[i].first.second;
    long long id = que[i].second;
    if (f1 < x) {
      for (long long j = f1; j < (long long)x; j++) {
        for (long long z = 0; z < (long long)fp[j].size(); z++) {
          int pri = fp[j][z];
          pos[pri].pop_front();
          if (!pos[pri].empty()) {
            up(pos[pri].front(), pri, pri - 1);
          }
        }
      }
      f1 = x;
    }
    ans.push_back({id, query(x, y, 1, 0, n)});
  }
  sort((ans).begin(), (ans).end());
  for (long long i = 0; i < (long long)ans.size(); i++) {
    cout << ans[i].second << '\n';
  }
  return 0;
}

```