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

<image>

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 = 1e5 + 5;
const long long MOD = 1e9 + 7;
int tol = 0;
int l, r;
long long arr[N];
long long phi[N];
long long mul(long long x, long long p) { return x > p ? x % p + p : x; }
long long eular(long long n) {
  long long ans = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans -= ans / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans -= ans / n;
  return ans;
}
long long quick_pow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans * a, mod);
    a = mul(a * a, mod);
    b >>= 1;
  }
  return ans;
}
long long fun(int pos, long long p) {
  if (pos == r || p == 1) {
    if (arr[pos] > p) return arr[pos] % p + p;
    return arr[pos];
  }
  return quick_pow(arr[pos], fun(pos + 1, phi[pos - l + 1]), p);
}
int main() {
  int n;
  long long p;
  scanf("%d%I64d", &n, &p);
  for (int i = (1); i <= (n); i++) scanf("%I64d", &arr[i]);
  phi[0] = p;
  while (phi[tol] != 1) {
    phi[tol + 1] = eular(phi[tol]);
    ++tol;
  }
  int m;
  scanf("%d", &m);
  for (int cas = (1); cas <= (m); cas++) {
    scanf("%d%d", &l, &r);
    long long ans = fun(l, p);
    printf("%I64d\n", ans - (ans >= phi[0] ? phi[0] : 0));
  }
  return 0;
}