#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
int n, m, tot, a[N], b[N], p[M / 10], f[M], pre[N], lst[M], ans[N];
bool flg[M];
struct Data {
  int l, r, id;
  bool operator<(const Data &rhs) const { return r < rhs.r; }
} q[N];
void sieve(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!flg[i]) p[++tot] = i, f[i] = i;
    for (int j = 1; j <= tot && i * p[j] <= n; ++j) {
      flg[i * p[j]] = 1, f[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
int pow(int x, int p) {
  int ret = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mod)
    if (p & 1) ret = 1LL * ret * x % mod;
  return ret;
}
int inv(int x) { return pow(x, mod - 2); }
void add(int x, int val) {
  for (; x <= n; x += x & -x) b[x] = 1LL * b[x] * val % mod;
}
int query(int x) {
  int ret = 1;
  for (; x; x ^= x & -x) ret = 1LL * ret * b[x] % mod;
  return ret;
}
void update(int i) {
  for (int x = a[i], p = f[x]; x > 1; p = f[x]) {
    add(i, p - 1), add(i, inv(p));
    if (lst[p]) add(lst[p], inv(p - 1)), add(lst[p], p);
    lst[p] = i;
    while (x % p == 0) x /= p;
  }
}
int main() {
  sieve(M - 5);
  scanf("%d", &n);
  pre[0] = 1;
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i]), pre[i] = 1LL * pre[i - 1] * a[i] % mod;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
  std::sort(q + 1, q + m + 1);
  for (int i = 0; i <= n; ++i) b[i] = 1;
  for (int i = 1, j = 0; i <= m; ++i) {
    int x = q[i].l, y = q[i].r;
    while (j < y) update(++j);
    ans[q[i].id] = 1LL * pre[y] * inv(pre[x - 1]) % mod * query(y) % mod *
                   inv(query(x - 1)) % mod;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 2e5 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
int n, m, tot, a[N], b[N], p[M / 10], f[M], pre[N], lst[M], ans[N];
bool flg[M];
struct Data {
  int l, r, id;
  bool operator<(const Data &rhs) const { return r < rhs.r; }
} q[N];
void sieve(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!flg[i]) p[++tot] = i, f[i] = i;
    for (int j = 1; j <= tot && i * p[j] <= n; ++j) {
      flg[i * p[j]] = 1, f[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
int pow(int x, int p) {
  int ret = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mod)
    if (p & 1) ret = 1LL * ret * x % mod;
  return ret;
}
int inv(int x) { return pow(x, mod - 2); }
void add(int x, int val) {
  for (; x <= n; x += x & -x) b[x] = 1LL * b[x] * val % mod;
}
int query(int x) {
  int ret = 1;
  for (; x; x ^= x & -x) ret = 1LL * ret * b[x] % mod;
  return ret;
}
void update(int i) {
  for (int x = a[i], p = f[x]; x > 1; p = f[x]) {
    add(i, p - 1), add(i, inv(p));
    if (lst[p]) add(lst[p], inv(p - 1)), add(lst[p], p);
    lst[p] = i;
    while (x % p == 0) x /= p;
  }
}
int main() {
  sieve(M - 5);
  scanf("%d", &n);
  pre[0] = 1;
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i]), pre[i] = 1LL * pre[i - 1] * a[i] % mod;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
  std::sort(q + 1, q + m + 1);
  for (int i = 0; i <= n; ++i) b[i] = 1;
  for (int i = 1, j = 0; i <= m; ++i) {
    int x = q[i].l, y = q[i].r;
    while (j < y) update(++j);
    ans[q[i].id] = 1LL * pre[y] * inv(pre[x - 1]) % mod * query(y) % mod *
                   inv(query(x - 1)) % mod;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```