Problem: Priests of the Quetzalcoatl cult want to build a tower to represent a power of their god. Tower is usually made of power-charged rocks. It is built with the help of rare magic by levitating the current top of tower and adding rocks at its bottom. If top, which is built from k - 1 rocks, possesses power p and we want to add the rock charged with power wk then value of power of a new tower will be {wk}p. 

Rocks are added from the last to the first. That is for sequence w1, ..., wm value of power will be

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After tower is built, its power may be extremely large. But still priests want to get some information about it, namely they want to know a number called cumulative power which is the true value of power taken modulo m. Priests have n rocks numbered from 1 to n. They ask you to calculate which value of cumulative power will the tower possess if they will build it from rocks numbered l, l + 1, ..., r. 

Input

First line of input contains two integers n (1 ≤ n ≤ 105) and m (1 ≤ m ≤ 109).

Second line of input contains n integers wk (1 ≤ wk ≤ 109) which is the power of rocks that priests have.

Third line of input contains single integer q (1 ≤ q ≤ 105) which is amount of queries from priests to you.

kth of next q lines contains two integers lk and rk (1 ≤ lk ≤ rk ≤ n). 

Output

Output q integers. k-th of them must be the amount of cumulative power the tower will have if is built from rocks lk, lk + 1, ..., rk.

Example

Input

6 1000000000
1 2 2 3 3 3
8
1 1
1 6
2 2
2 3
2 4
4 4
4 5
4 6


Output

1
1
2
4
256
3
27
597484987

Note

327 = 7625597484987

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
long long m, a[N], phi[100];
long long POW(long long a, long long b, long long md) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      if (ans * a >= md)
        ans = ans * a % md + md;
      else
        ans = ans * a;
    }
    b >>= 1;
    if (a * a >= md)
      a = a * a % md + md;
    else
      a = a * a;
  }
  return ans;
}
long long PHI(long long k) {
  long long ans = k;
  for (long long p = 2; p * p <= k; p++) {
    if (k % p == 0) {
      while (k % p == 0) k /= p;
      ans -= (ans / p);
    }
  }
  if (k > 1) {
    ans -= (ans / k);
  }
  return ans;
}
int main() {
  int n;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  phi[0] = m;
  int idx = 0;
  for (int i = 1; i < 100; i++) {
    phi[i] = PHI(phi[i - 1]);
    if (phi[i - 1] == 1) {
      idx = i;
      break;
    }
    if (phi[i] == 1) {
      idx = i + 1;
      break;
    }
  }
  idx--;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    long long ans = 1LL;
    for (int i = min(r, l + idx); i >= l; i--) {
      ans = POW(a[i], ans, phi[i - l]);
    }
    cout << ans % m << "\n";
  }
  return 0;
}