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;
int n, l, r, num, p;
int m, w[100010], phi[300];
int count_phi(int p) {
  int l = 1, i = 2;
  if (p == 1) return 0;
  while (p != 1) {
    while (p % i != 0) i++;
    l *= (i - 1);
    p /= i;
    while (p % i == 0) {
      p /= i;
      l *= i;
    }
  }
  return l;
}
int mpow(int a, int n, int mod) {
  if (n == 0) return 1;
  int b = mpow(a, n / 2, mod);
  b = (1LL * b * b) % mod;
  if (n % 2 == 1) b = (1LL * a * b) % mod;
  return b;
}
int com(int a, int b, int c) {
  long long d = 1;
  if ((a == 1) && (c > 1)) return 1;
  for (int i = 1; i <= b; i++) {
    if (d >= c / a) return 0;
    d = d * a;
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  phi[0] = m;
  while (phi[num++] != 0) phi[num] = count_phi(phi[num - 1]);
  num -= 2;
  scanf("%d", &p);
  for (int i = 0; i < p; i++) {
    scanf("%d %d", &l, &r);
    int k = 1;
    for (int i = min(r, l + num); i >= l; i--) {
      if (com(w[i], k, phi[i - l]))
        k = mpow(w[i], k, phi[i - l]);
      else
        k = mpow(w[i], k, phi[i - l]) + phi[i - l];
    }
    printf("%d\n", k % m);
  }
}