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 getmod(long long x, int mod) {
  if (x >= 32) return 32 + (x - 32) % mod;
  return x;
}
int phi(int n) {
  int res = n;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      res -= res / i;
    }
  if (n > 1) res -= res / n;
  return res;
}
vector<int> G;
int bp(int a, int n, int mod) {
  if (n == 0) return 1;
  if (n % 2) return getmod(1ll * bp(a, n - 1, mod) * a, mod);
  int t = bp(a, n / 2, mod);
  return getmod(1ll * t * t, mod);
}
int main() {
  int n, m;
  cin >> n >> m;
  G.push_back(m);
  while (m > 1) m = phi(m), G.push_back(m);
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    r = min(r, l + (int)G.size() - 1);
    int ans = 1;
    for (int j = r; j >= l; j--) {
      int x = ans;
      ans = bp(a[j], x, G[j - l]);
    }
    cout << ans % G[0] << "\n";
  }
  return 0;
}