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, nT, t[1 << 17], w[1 << 17], Q;
int l, r;
long long mul(long long a, long long b, long long m) {
  return a * b >= m ? a * b % m + m : a * b;
}
long long mpow(long long b, long long exp, long long m) {
  long long r = 1;
  while (exp) {
    if (exp & 1) r = mul(r, b, m);
    b = mul(b, b, m);
    exp /= 2;
  }
  return r;
}
int phi(int n) {
  int r = n;
  for (int i = 2; i <= n; ++i) {
    if (n % i) continue;
    r = (r / i) * (i - 1);
    while (!(n % i)) n /= i;
  }
  return r;
}
int main() {
  cin >> N >> t[0];
  for (int i = 0; i++ < N;) cin >> w[i];
  for (nT = 0; t[nT] > 1; t[nT + 1] = phi(t[nT]), ++nT)
    ;
  cin >> Q;
  while (Q--) {
    cin >> l >> r;
    long long v = 1;
    for (int i = min(nT, r - l); ~i; --i) v = mpow(w[i + l], v, t[i]);
    cout << v % t[0] << "\n";
  }
}
