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 MAXN = 1e5 + 5;
int w[MAXN];
map<int, int> f;
long long Calc(long long Res, long long Mod) {
  if (Res > Mod) return Res % Mod + Mod;
  return Res;
}
long long power(long long a, long long n, long long Mod) {
  long long Res = 1;
  while (n) {
    if (n & 1) Res = Calc(Res * a, Mod);
    a = Calc(a * a, Mod);
    n >>= 1;
  }
  return Res;
}
int phi(int n) {
  int result = 1;
  if (f.count(n)) return f[n];
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      while (n % i == 0) {
        n /= i;
        result *= i;
      }
      result /= i;
      result *= (i - 1);
    }
  }
  if (n != 1) result *= (n - 1);
  f[n] = result;
  return f[n];
}
long long Solve(int l, int r, int Mod) {
  if (l == r || Mod == 1) return Calc(w[l], Mod);
  return power(w[l], Solve(l + 1, r, phi(Mod)), Mod);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q, n, Mod;
  cin >> n >> Mod;
  for (int i = 1; i <= n; ++i) cin >> w[i];
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    long long ans = Solve(l, r, Mod);
    cout << ans % Mod << endl;
  }
  return 0;
}
