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;
int n, q, flag;
int w[N];
int phi[100];
int calc_phi(int x);
int solve(int l, int r, int id);
int pow(int x, int k, int p);
int main() {
  scanf("%d%d", &n, &phi[0]);
  for (int i = 0; phi[i] != 1; ++i) phi[i + 1] = calc_phi(phi[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);
  scanf("%d", &q);
  while (q--) {
    int l, r;
    flag = 0;
    scanf("%d%d", &l, &r);
    printf("%d\n", solve(l, r, 0));
  }
  return 0;
}
int pow(int x, int k, int p) {
  int ret = 1;
  bool f = false;
  flag = 0;
  if (x >= p) {
    flag = 1;
    f = true;
  }
  x %= p;
  while (k) {
    if (k & 1) {
      if ((long long)ret * x >= p || f) flag = 1;
      ret = (long long)ret * x % p;
    }
    k >>= 1;
    if ((long long)x * x >= p) f = true;
    x = (long long)x * x % p;
  }
  return ret;
}
int solve(int l, int r, int id) {
  if (w[l] % phi[id] == 0) {
    flag = 1;
    return 0;
  }
  if (l == r) {
    if (w[l] >= phi[id]) flag = 1;
    return w[l] % phi[id];
  }
  int q = solve(l + 1, r, id + 1);
  return pow(w[l], q + phi[id + 1] * flag, phi[id]);
}
int calc_phi(int x) {
  int ans = x, sie = x;
  for (int i = 2; i * i <= x; ++i)
    if (sie % i == 0) {
      ans = ans / i * (i - 1);
      while (sie % i == 0) sie /= i;
    }
  if (sie != 1) ans = ans / sie * (sie - 1);
  return ans;
}