#include <bits/stdc++.h>
using namespace std;
long long bigMod(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % 1000000007;
      y--;
    } else {
      x = (x * x) % 1000000007;
      y >>= 1;
    }
  }
  return res;
}
struct query {
  int l, r, id;
} Q[200005];
const int k = 450;
int ara[200005], freq[1000006];
bool cmp(query &a, query &b) {
  int b1 = a.l / k, b2 = b.l / k;
  if (b1 == b2) return a.r < b.r;
  return b1 < b2;
}
int l = 0, r = -1, ans[200005];
long long sum, res;
vector<int> fac[200005];
void Add(int x) {
  sum = (sum * ara[x]) % 1000000007;
  for (auto &y : fac[x]) {
    if (!freq[y]) {
      sum = (sum * (y - 1)) % 1000000007;
      res = (res * y) % 1000000007;
    }
    freq[y]++;
  }
}
void Del(int x) {
  res = (res * ara[x]) % 1000000007;
  for (auto &y : fac[x]) {
    freq[y]--;
    if (!freq[y]) {
      res = (res * (y - 1)) % 1000000007;
      sum = (sum * y) % 1000000007;
    }
  }
}
vector<int> prime;
bitset<1000006> a;
void sieve() {
  a[0] = a[1] = 1;
  for (int i = 4; i < 200005; i += 2) a[i] = 1;
  for (int i = 3; i * i < 200005; i++) {
    if (!a[i]) {
      for (int j = i * i; j < 200005; j += 2 * i) a[j] = 1;
    }
  }
  prime.push_back(2);
  for (int i = 3; i < 200005; i += 2)
    if (!a[i]) prime.push_back(i);
}
void primefact(int n) {
  int x = n;
  n = ara[n];
  for (int i = 0; prime[i] * prime[i] <= n && i < int(prime.size()); i++) {
    if (n % prime[i] == 0) {
      while (n % prime[i] == 0) {
        n /= prime[i];
      }
      fac[x].push_back(prime[i]);
    }
  }
  if (n > 1) {
    fac[x].push_back(n);
  }
}
int main() {
  sieve();
  int q, n;
  scanf("%d", &n);
  for (int i = int(0); i < int(n); i++) {
    scanf("%d", &ara[i]);
    primefact(i);
  }
  scanf("%d", &q);
  for (int i = int(0); i < int(q); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    Q[i].l = x, Q[i].r = y;
    Q[i].id = i;
  }
  sum = 1;
  sort(Q, Q + q, cmp);
  for (int i = int(0); i < int(q); i++) {
    res = 1;
    while (l > Q[i].l) Add(--l);
    while (r < Q[i].r) Add(++r);
    while (l < Q[i].l) Del(l++);
    while (r > Q[i].r) Del(r--);
    sum = (sum * bigMod(res, 1000000007 - 2)) % 1000000007;
    ans[Q[i].id] = sum;
  }
  for (int i = int(0); i < int(q); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 bigMod(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % 1000000007;
      y--;
    } else {
      x = (x * x) % 1000000007;
      y >>= 1;
    }
  }
  return res;
}
struct query {
  int l, r, id;
} Q[200005];
const int k = 450;
int ara[200005], freq[1000006];
bool cmp(query &a, query &b) {
  int b1 = a.l / k, b2 = b.l / k;
  if (b1 == b2) return a.r < b.r;
  return b1 < b2;
}
int l = 0, r = -1, ans[200005];
long long sum, res;
vector<int> fac[200005];
void Add(int x) {
  sum = (sum * ara[x]) % 1000000007;
  for (auto &y : fac[x]) {
    if (!freq[y]) {
      sum = (sum * (y - 1)) % 1000000007;
      res = (res * y) % 1000000007;
    }
    freq[y]++;
  }
}
void Del(int x) {
  res = (res * ara[x]) % 1000000007;
  for (auto &y : fac[x]) {
    freq[y]--;
    if (!freq[y]) {
      res = (res * (y - 1)) % 1000000007;
      sum = (sum * y) % 1000000007;
    }
  }
}
vector<int> prime;
bitset<1000006> a;
void sieve() {
  a[0] = a[1] = 1;
  for (int i = 4; i < 200005; i += 2) a[i] = 1;
  for (int i = 3; i * i < 200005; i++) {
    if (!a[i]) {
      for (int j = i * i; j < 200005; j += 2 * i) a[j] = 1;
    }
  }
  prime.push_back(2);
  for (int i = 3; i < 200005; i += 2)
    if (!a[i]) prime.push_back(i);
}
void primefact(int n) {
  int x = n;
  n = ara[n];
  for (int i = 0; prime[i] * prime[i] <= n && i < int(prime.size()); i++) {
    if (n % prime[i] == 0) {
      while (n % prime[i] == 0) {
        n /= prime[i];
      }
      fac[x].push_back(prime[i]);
    }
  }
  if (n > 1) {
    fac[x].push_back(n);
  }
}
int main() {
  sieve();
  int q, n;
  scanf("%d", &n);
  for (int i = int(0); i < int(n); i++) {
    scanf("%d", &ara[i]);
    primefact(i);
  }
  scanf("%d", &q);
  for (int i = int(0); i < int(q); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    Q[i].l = x, Q[i].r = y;
    Q[i].id = i;
  }
  sum = 1;
  sort(Q, Q + q, cmp);
  for (int i = int(0); i < int(q); i++) {
    res = 1;
    while (l > Q[i].l) Add(--l);
    while (r < Q[i].r) Add(++r);
    while (l < Q[i].l) Del(l++);
    while (r > Q[i].r) Del(r--);
    sum = (sum * bigMod(res, 1000000007 - 2)) % 1000000007;
    ans[Q[i].id] = sum;
  }
  for (int i = int(0); i < int(q); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```