#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int const nmax = 200000;
int const valmax = 1000000;
int const modulo = 1000000007;
void gcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, x, y);
    int aux = x;
    x = y;
    y = aux - a / b * y;
  }
}
int invmod(int a, int modulo) {
  int x, y;
  gcd(a, modulo, x, y);
  x %= modulo;
  if (x < 0) x += modulo;
  return x;
}
int v[1 + nmax];
bitset<1 + valmax> ciur;
vector<int> primes;
void computeprimes() {
  ciur[1] = 1;
  for (int i = 2; i * i <= valmax; i++) {
    if (ciur[i] == 0)
      for (int j = i * i; j <= valmax; j += i) ciur[j] = 1;
  }
  for (int i = 1; i <= valmax; i++)
    if (ciur[i] == 0) primes.push_back(i);
}
vector<pair<int, int>> query[1 + nmax];
int last[1 + valmax];
class SegmentTree {
 private:
  vector<int> aint;

 public:
  SegmentTree(int n) { aint.resize(4 * n); }
  void build(int node, int from, int to) {
    aint[node] = 1;
    if (from < to) {
      int mid = (from + to) / 2;
      build(node * 2, from, mid);
      build(node * 2 + 1, mid + 1, to);
    }
  }
  void update(int node, int from, int to, int x, int y, int val) {
    if (from == x && to == y)
      aint[node] = 1LL * aint[node] * val % modulo;
    else {
      int mid = (from + to) / 2;
      if (x <= mid)
        update(node * 2, from, mid, x, (((mid) < (y)) ? (mid) : (y)), val);
      if (mid + 1 <= y)
        update(node * 2 + 1, mid + 1, to, (((mid + 1) < (x)) ? (x) : (mid + 1)),
               y, val);
    }
  }
  int query(int node, int from, int to, int x) {
    if (from < to) {
      int mid = (from + to) / 2;
      int result = aint[node];
      if (x <= mid)
        return 1LL * result * query(node * 2, from, mid, x) % modulo;
      else
        return 1LL * result * query(node * 2 + 1, mid + 1, to, x) % modulo;
    } else
      return aint[node];
  }
};
void updateprime(int pos, int val, int n, SegmentTree &aint) {
  int coef = (modulo + 1LL - invmod(val, modulo)) % modulo;
  aint.update(1, 1, n, pos, last[val] - 1, coef);
  last[val] = pos;
}
void update(int pos, int n, SegmentTree &aint) {
  for (int i = 0; i < primes.size() && primes[i] * primes[i] <= v[pos]; i++) {
    if (v[pos] % primes[i] == 0) {
      updateprime(pos, primes[i], n, aint);
      while (v[pos] % primes[i] == 0) v[pos] /= primes[i];
    }
  }
  if (1 < v[pos]) updateprime(pos, v[pos], n, aint);
}
int sol[1 + nmax], pref[1 + nmax];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n;
  computeprimes();
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= valmax; i++) last[i] = n + 1;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    query[x].push_back({i, y});
  }
  pref[0] = 1;
  for (int i = 1; i <= n; i++) pref[i] = 1LL * pref[i - 1] * v[i] % modulo;
  SegmentTree aint(n);
  aint.build(1, 1, n);
  for (int i = n; 1 <= i; i--) {
    update(i, n, aint);
    for (int h = 0; h < query[i].size(); h++) {
      int y = query[i][h].second;
      sol[query[i][h].first] = 1LL * pref[y] * invmod(pref[i - 1], modulo) %
                               modulo * aint.query(1, 1, n, y) % modulo;
    }
  }
  for (int i = 1; i <= m; i++) cout << sol[i] << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp 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;
using ll = long long;
int const nmax = 200000;
int const valmax = 1000000;
int const modulo = 1000000007;
void gcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, x, y);
    int aux = x;
    x = y;
    y = aux - a / b * y;
  }
}
int invmod(int a, int modulo) {
  int x, y;
  gcd(a, modulo, x, y);
  x %= modulo;
  if (x < 0) x += modulo;
  return x;
}
int v[1 + nmax];
bitset<1 + valmax> ciur;
vector<int> primes;
void computeprimes() {
  ciur[1] = 1;
  for (int i = 2; i * i <= valmax; i++) {
    if (ciur[i] == 0)
      for (int j = i * i; j <= valmax; j += i) ciur[j] = 1;
  }
  for (int i = 1; i <= valmax; i++)
    if (ciur[i] == 0) primes.push_back(i);
}
vector<pair<int, int>> query[1 + nmax];
int last[1 + valmax];
class SegmentTree {
 private:
  vector<int> aint;

 public:
  SegmentTree(int n) { aint.resize(4 * n); }
  void build(int node, int from, int to) {
    aint[node] = 1;
    if (from < to) {
      int mid = (from + to) / 2;
      build(node * 2, from, mid);
      build(node * 2 + 1, mid + 1, to);
    }
  }
  void update(int node, int from, int to, int x, int y, int val) {
    if (from == x && to == y)
      aint[node] = 1LL * aint[node] * val % modulo;
    else {
      int mid = (from + to) / 2;
      if (x <= mid)
        update(node * 2, from, mid, x, (((mid) < (y)) ? (mid) : (y)), val);
      if (mid + 1 <= y)
        update(node * 2 + 1, mid + 1, to, (((mid + 1) < (x)) ? (x) : (mid + 1)),
               y, val);
    }
  }
  int query(int node, int from, int to, int x) {
    if (from < to) {
      int mid = (from + to) / 2;
      int result = aint[node];
      if (x <= mid)
        return 1LL * result * query(node * 2, from, mid, x) % modulo;
      else
        return 1LL * result * query(node * 2 + 1, mid + 1, to, x) % modulo;
    } else
      return aint[node];
  }
};
void updateprime(int pos, int val, int n, SegmentTree &aint) {
  int coef = (modulo + 1LL - invmod(val, modulo)) % modulo;
  aint.update(1, 1, n, pos, last[val] - 1, coef);
  last[val] = pos;
}
void update(int pos, int n, SegmentTree &aint) {
  for (int i = 0; i < primes.size() && primes[i] * primes[i] <= v[pos]; i++) {
    if (v[pos] % primes[i] == 0) {
      updateprime(pos, primes[i], n, aint);
      while (v[pos] % primes[i] == 0) v[pos] /= primes[i];
    }
  }
  if (1 < v[pos]) updateprime(pos, v[pos], n, aint);
}
int sol[1 + nmax], pref[1 + nmax];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n;
  computeprimes();
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= valmax; i++) last[i] = n + 1;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    query[x].push_back({i, y});
  }
  pref[0] = 1;
  for (int i = 1; i <= n; i++) pref[i] = 1LL * pref[i - 1] * v[i] % modulo;
  SegmentTree aint(n);
  aint.build(1, 1, n);
  for (int i = n; 1 <= i; i--) {
    update(i, n, aint);
    for (int h = 0; h < query[i].size(); h++) {
      int y = query[i][h].second;
      sol[query[i][h].first] = 1LL * pref[y] * invmod(pref[i - 1], modulo) %
                               modulo * aint.query(1, 1, n, y) % modulo;
    }
  }
  for (int i = 1; i <= m; i++) cout << sol[i] << '\n';
  return 0;
}

```