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
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 50, M = 105;
long long n, m, q, w[N], pn, ps[M];
long long phi(long long x) {
  long long res = 1;
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      res *= i - 1;
      x /= i;
      for (; x % i == 0; x /= i) res *= i;
    }
  if (x > 1) res *= x - 1;
  return res;
}
long long pw(long long a, long long k, long long p) {
  long long res = 1;
  for (; k; a = (a * a < p ? a * a : (a * a % p + p)), k >>= 1) {
    if (k & 1) res = (res * a < p ? res * a : (res * a % p + p));
  }
  return res;
}
long long sol(long long l, long long r, long long pos) {
  if (l == r)
    return (w[l] < ps[pos] ? w[l] : (w[l] % ps[pos] + ps[pos]));
  else if (ps[pos] == 1)
    return 1;
  return pw(w[l], sol(l + 1, r, pos + 1), ps[pos]);
}
int main(int argc, char *argv[]) {
  scanf("%lld%lld", &n, &m);
  for (ps[++pn] = m; ps[pn] != 1; ++pn) ps[pn + 1] = phi(ps[pn]);
  for (long long i = 1; i <= (n); ++i) scanf("%lld", w + i);
  scanf("%lld", &q);
  ;
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    ;
    printf("%lld\n", sol(l, r, 1) % m);
  }
}
