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;
map<long long, long long> mp;
long long phi(long long n) {
  if (mp[n]) return mp[n];
  long long k = n;
  long long ans = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      ans -= ans / i;
    }
  }
  if (n > 1) {
    ans -= ans / n;
  }
  return mp[k] = ans;
}
long long md(long long a, long long b) { return (a < b ? a : a % b + b); }
long long modpow(long long n, long long k, long long M) {
  if (k == 0) return md(1, M);
  long long tmp = modpow(n, k / 2, M);
  tmp = md(tmp * tmp, M);
  if (k % 2)
    return md(tmp * n, M);
  else
    return tmp;
}
long long solve(int l, int r, long long M, long long a[]) {
  if (r == l || M == 1) return md(a[l], M);
  return modpow(a[l], solve(l + 1, r, phi(M), a), M);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long M;
  cin >> n >> M;
  long long a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r, M, a) % M << '\n';
  }
  return 0;
}