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;
long long n, m, T, x, y, a[101000];
map<long long, long long> phi;
long long GP(long long x) {
  if (phi.count(x)) return phi[x];
  long long res = x, z = x;
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      for (; x % i == 0;) x /= i;
      res = res / i * (i - 1);
    }
  if (x > 1) res = res / x * (x - 1);
  return phi[z] = res;
}
long long K(long long x, long long y, long long mod) {
  long long res = 1, fl = pow(x + 0., y) > mod - 0.5;
  for (; y; y >>= 1, x = x * x % mod)
    if (y & 1) res = res * x % mod;
  return res + fl * mod;
}
long long que(int l, int r, long long mod) {
  if (l == r) return a[l];
  if (mod == 1) return 1;
  return K(a[l], que(l + 1, r, GP(mod)), mod);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  for (cin >> T; T--;) {
    scanf("%I64d%I64d", &x, &y);
    printf("%I64d\n", que(x, y, m) % m);
  }
}
