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 maxn = 1e5 + 5;
int n, m;
int a[maxn];
long long _phi(long long x) {
  long long ans = 1;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans * (i - 1);
      x /= i;
      while (x % i == 0) {
        x /= i;
        ans = ans * i;
      }
    }
  }
  if (x > 1) ans = ans * (x - 1);
  return ans;
}
vector<int> phi;
long long pw(long long x, long long k, long long p) {
  long long ans = 1;
  for (; k; k >>= 1) {
    if (k & 1) ans = ans * x % p;
    x = x * x % p;
  }
  return ans;
}
long long mode(long long a, long long b) {
  if (a < b) return a;
  return a % b + b;
}
long long calc(int l, int r, int k) {
  if (l > r || k >= phi.size()) return 1;
  long long t = calc(l + 1, r, k + 1), ans = 1, x = a[l];
  for (; t; t >>= 1) {
    if (t & 1) ans = mode(ans * x, phi[k]);
    x = mode(x * x, phi[k]);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  long long mm = m;
  while (m > 1) {
    phi.push_back(m);
    m = _phi(m);
  }
  m = mm;
  phi.push_back(1);
  phi.push_back(1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", calc(l, r, 0) % m);
  }
  return 0;
}