#include <bits/stdc++.h>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.0000000001;
using namespace std;
int tab[1000005], W[1000005], O[1000005], F[1000005], S[1000005];
vector<int> I[200005];
void sito(int n) {
  for (int i = 2; i * i <= n; i++)
    if (!tab[i])
      for (int j = i * i; j <= n; j += i) tab[j] = i;
}
struct Zap {
  int p, k, nr;
};
Zap Z[200005];
int T[200005];
int pier;
bool cmp(Zap a, Zap b) {
  if (a.p / pier != b.p / pier) return a.p / pier < b.p / pier;
  if (a.p / pier % 2 == 0) return a.k < b.k;
  return a.k > b.k;
}
int A[200005];
int Pow(int a, int b) {
  int wynik = 1;
  while (b) {
    if (b & 1) wynik = ((long long int)wynik * a) % mod;
    a = ((long long int)a * a) % mod;
    b /= 2;
  }
  return wynik;
}
int wynik = 1;
void dodaj(int x) {
  if (x == 0) return;
  wynik = ((long long int)wynik * T[x]) % mod;
  for (auto v : I[x]) {
    if (!W[v]) wynik = ((long long int)wynik * F[v]) % mod;
    W[v]++;
  }
}
void odejmij(int x) {
  if (x == 0) return;
  wynik = ((long long int)wynik * O[T[x]]) % mod;
  for (auto v : I[x]) {
    W[v]--;
    if (!W[v]) wynik = ((long long int)wynik * S[v]) % mod;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  sito(1000000);
  for (int i = 1; i <= 1000000; i++)
    if (!tab[i]) tab[i] = i;
  for (int i = 1; i <= 1000000; i++) {
    O[i] = Pow(i, mod - 2);
    F[i] = ((long long int)(i - 1) * O[i]) % mod;
    S[i] = ((long long int)i * (O[i - 1])) % mod;
  }
  for (int i = 1; i <= n; i++) {
    cin >> T[i];
    int x = T[i];
    int last = -1;
    while (x != 1) {
      if (last != tab[x]) I[i].push_back(tab[x]);
      last = tab[x];
      x /= tab[x];
    }
  }
  int q;
  cin >> q;
  pier = sqrt(q + n);
  for (int j = 1; j <= q; j++) {
    int a, b;
    cin >> a >> b;
    Z[j] = {a, b, j};
  }
  sort(Z + 1, Z + q + 1, cmp);
  int p = 0, k = 0;
  for (int i = 1; i <= q; i++) {
    while (p > Z[i].p) dodaj(--p);
    while (k < Z[i].k) dodaj(++k);
    while (p < Z[i].p) odejmij(p++);
    while (k > Z[i].k) odejmij(k--);
    A[Z[i].nr] = wynik;
  }
  for (int i = 1; i <= q; i++) cout << A[i] << "\n";
  return 0;
}

### Prompt

Generate 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>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.0000000001;
using namespace std;
int tab[1000005], W[1000005], O[1000005], F[1000005], S[1000005];
vector<int> I[200005];
void sito(int n) {
  for (int i = 2; i * i <= n; i++)
    if (!tab[i])
      for (int j = i * i; j <= n; j += i) tab[j] = i;
}
struct Zap {
  int p, k, nr;
};
Zap Z[200005];
int T[200005];
int pier;
bool cmp(Zap a, Zap b) {
  if (a.p / pier != b.p / pier) return a.p / pier < b.p / pier;
  if (a.p / pier % 2 == 0) return a.k < b.k;
  return a.k > b.k;
}
int A[200005];
int Pow(int a, int b) {
  int wynik = 1;
  while (b) {
    if (b & 1) wynik = ((long long int)wynik * a) % mod;
    a = ((long long int)a * a) % mod;
    b /= 2;
  }
  return wynik;
}
int wynik = 1;
void dodaj(int x) {
  if (x == 0) return;
  wynik = ((long long int)wynik * T[x]) % mod;
  for (auto v : I[x]) {
    if (!W[v]) wynik = ((long long int)wynik * F[v]) % mod;
    W[v]++;
  }
}
void odejmij(int x) {
  if (x == 0) return;
  wynik = ((long long int)wynik * O[T[x]]) % mod;
  for (auto v : I[x]) {
    W[v]--;
    if (!W[v]) wynik = ((long long int)wynik * S[v]) % mod;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  sito(1000000);
  for (int i = 1; i <= 1000000; i++)
    if (!tab[i]) tab[i] = i;
  for (int i = 1; i <= 1000000; i++) {
    O[i] = Pow(i, mod - 2);
    F[i] = ((long long int)(i - 1) * O[i]) % mod;
    S[i] = ((long long int)i * (O[i - 1])) % mod;
  }
  for (int i = 1; i <= n; i++) {
    cin >> T[i];
    int x = T[i];
    int last = -1;
    while (x != 1) {
      if (last != tab[x]) I[i].push_back(tab[x]);
      last = tab[x];
      x /= tab[x];
    }
  }
  int q;
  cin >> q;
  pier = sqrt(q + n);
  for (int j = 1; j <= q; j++) {
    int a, b;
    cin >> a >> b;
    Z[j] = {a, b, j};
  }
  sort(Z + 1, Z + q + 1, cmp);
  int p = 0, k = 0;
  for (int i = 1; i <= q; i++) {
    while (p > Z[i].p) dodaj(--p);
    while (k < Z[i].k) dodaj(++k);
    while (p < Z[i].p) odejmij(p++);
    while (k > Z[i].k) odejmij(k--);
    A[Z[i].nr] = wynik;
  }
  for (int i = 1; i <= q; i++) cout << A[i] << "\n";
  return 0;
}

```