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;
int a[100005], phi[100], n, m, mod;
int get(int x) {
  int ans = x;
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) ans = ans / x * (x - 1);
  return ans;
}
int gg(long long x, int p) { return x >= p ? x % p + p : x; }
int qpow(int a, int b, int p) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = gg(1ll * ans * a, p);
    a = gg(1ll * a * a, p);
    b >>= 1;
  }
  return ans;
}
int dfs(int l, int r, int i) {
  if (l == r || phi[i] == 1) return gg(a[l], phi[i]);
  return qpow(a[l], dfs(l + 1, r, i + 1), phi[i]);
}
int main() {
  scanf("%d%d", &n, &mod);
  phi[0] = mod;
  for (int i = 1; i <= 30; i++) {
    phi[i] = get(phi[i - 1]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", dfs(l, r, 0) % mod);
  }
  return 0;
}