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;
long long a[1000005];
map<int, int> phi;
int MOD(long long x, int p) { return x > p ? x % p + p : x; }
long long poww(long long a, int b, int m) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res = MOD(res, m);
    a *= a, a = MOD(a, m);
    b >>= 1;
  }
  return res;
}
int getphi(int m) {
  if (phi[m] != 0) return phi[m];
  int res = m;
  for (int i = 2; i <= m / i; i++) {
    if (m % i == 0) {
      res = res / i * (i - 1);
      while (m % i == 0) {
        m = m / i;
      }
    }
  }
  if (m > 1) res = res / m * (m - 1);
  return phi[m] = res;
}
long long solve(int L, int R, int m) {
  if (m == 1) return 1;
  if (L == R) return MOD(a[R], m);
  return poww(a[L], solve(L + 1, R, getphi(m)), m);
}
int main() {
  int n;
  long long m;
  int q;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> q;
  while (q--) {
    int L;
    int R;
    cin >> L >> R;
    cout << solve(L, R, m) % m << "\n";
  }
  return 0;
}