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>
inline int Read() {
  char c = getchar();
  int n = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9')
    n = (n << 1) + (n << 3) + (c ^ 48), c = getchar();
  return n;
}
int n, m, q, t, eul[100];
long long a[100005];
inline int Euler(int k) {
  int res = k;
  for (int i = 2; i * i <= k; i++)
    if (k % i == 0) {
      res = res / i * (i - 1);
      while (k % i == 0) k = k / i;
    }
  return k == 1 ? res : res / k * (k - 1);
}
inline int Pow(long long k, int d, int p) {
  long long res = 1;
  while (d > 0) {
    if (d & 1) res = ((res * k) < (p) ? (res * k) : (res * k) % (p) + (p));
    k = ((k * k) < (p) ? (k * k) : (k * k) % (p) + (p));
    d >>= 1;
  }
  return res;
}
int Solve(int d, int l, int r, int p) {
  if (l == r || p == 1) return ((a[l]) < (p) ? (a[l]) : (a[l]) % (p) + (p));
  int ans = Solve(d + 1, l + 1, r, eul[d]);
  return Pow(a[l], ans, p);
}
int main() {
  n = Read(), m = Read();
  eul[0] = m;
  for (t = 1; eul[t - 1] != 1; t++) eul[t] = Euler(eul[t - 1]);
  for (int i = 1; i <= n; i++) a[i] = Read();
  q = Read();
  for (int i = 1; i <= q; i++) {
    int l = Read(), r = Read();
    printf("%d\n", Solve(1, l, r, m) % m);
  }
  return 0;
}