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
#include <bits/stdc++.h>
using namespace std;
int N, Q, i, L, R, P[1000];
long long A[100000];
int euc(int X) {
  int i, R = X;
  for (i = 2; i * i <= R; i++) {
    if (R % i == 0) {
      for (X -= X / i; R % i == 0; R /= i) {
      }
    }
  }
  if (R > 1) {
    X -= X / R;
  }
  return X;
}
long long kal(long long X, long long Y, int M) {
  return min((X * Y) % M + M, X * Y);
}
long long pan(long long X, int Y, int M) {
  if (Y == 0) {
    return 1;
  }
  long long R = pan(X, Y >> 1, M);
  R = kal(R, R, M);
  if (Y & 1) {
    R = kal(R, X, M);
  }
  return R;
}
long long ans(int D) {
  if (D == R) {
    return pan(A[D], 1, P[D - L]);
  }
  if (P[D - L] == 1) {
    return min(A[D], (long long)2);
  }
  return pan(A[D], ans(D + 1), P[D - L]);
  return R;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> P[0];
  for (i = 0; i < N; i++) {
    cin >> A[i];
  }
  for (i = 1; P[i - 1] > 1; i++) {
    P[i] = euc(P[i - 1]);
  }
  cin >> Q;
  for (i = 0; i < Q; i++) {
    cin >> L >> R;
    L--;
    R--;
    cout << ans(L) % P[0] << '\n';
  }
}
