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 N = 1e5 + 10;
int n, q;
long long a[N];
long long m;
map<long long, long long> f;
long long phi(long long n) {
  if (f.count(n)) return f[n];
  long long ans = n, z = n;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      ans -= ans / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans -= ans / n;
  return f[z] = ans;
}
long long Pow(long long a, long long b, long long mod) {
  long long ret = 1;
  long long fl = a >= mod;
  for (; b; b >>= 1) {
    if (b & 1) {
      ret *= a;
      if (ret >= mod) fl = 1, ret %= mod;
    }
    a *= a;
    if (a >= mod) a %= mod, fl = 1;
  }
  return ret + fl * mod;
}
long long solve(int l, int r, long long mod) {
  if (l == r) return a[l];
  if (mod == 1) return 1;
  return Pow(a[l], solve(l + 1, r, phi(mod)), mod);
}
int main() {
  scanf("%d%lld", &n, &m);
  for (int i(1); i <= (n); ++i) scanf("%lld", a + i);
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    printf("%lld\n", solve(x, y, m) % m);
  }
  return 0;
}