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>
using namespace std;
const int ONE = 1000005;
int get() {
  int res = 1, Q = 1;
  char c;
  while ((c = getchar()) < 48 || c > 57)
    if (c == '-') Q = -1;
  if (Q) res = c - 48;
  while ((c = getchar()) >= 48 && c <= 57) res = res * 10 + c - 48;
  return res * Q;
}
int Getphi(int x) {
  long long res = x;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      res = res * (i - 1) / i;
      while (x % i == 0) x /= i;
    }
  if (x > 1) res = res * (x - 1) / x;
  return res;
}
int n, p, a[ONE];
int phi[ONE], p_num;
int Quickpow(int a, int b, int MOD) {
  int res = 1;
  while (b) {
    if (b & 1)
      res = (long long)res * a >= MOD ? (long long)res * a % MOD + MOD
                                      : (long long)res * a;
    ;
    a = (long long)a * a >= MOD ? (long long)a * a % MOD + MOD
                                : (long long)a * a;
    ;
    b >>= 1;
  }
  return res;
}
int Solve(int l, int r, int T) {
  int x = a[l];
  if (l == r || T == p_num) return x >= phi[T] ? x % phi[T] + phi[T] : x;
  ;
  return Quickpow(x, Solve(l + 1, r, T + 1), phi[T]);
}
int main() {
  n = get(), phi[0] = p = get();
  while (p > 1) phi[++p_num] = p = Getphi(p);
  for (int i = 1; i <= n; i++) a[i] = get();
  int Q = get();
  while (Q--) {
    int l = get(), r = get();
    printf("%d\n", Solve(l, r, 0) % phi[0]);
  }
}