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 maxn = 1e5 + 1;
int n, m, q;
int w[maxn];
map<int, int> pre;
int phi(int n) {
  int x = n;
  if (pre.find(x) != pre.end()) {
    return pre[x];
  }
  int ans = x;
  int sx = (int)sqrt(x) + 1;
  for (int i = 2; i < sx; ++i) {
    if (x % i == 0) {
      while (x % i == 0) {
        x /= i;
      }
      ans -= ans / i;
    }
  }
  if (x > 1) {
    ans -= ans / x;
  }
  return pre[n] = ans;
}
int MOD(long long a, long long b) {
  if (a < b) {
    return a;
  }
  return a % b + b;
}
int power(int x, int y, int mod) {
  int ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = MOD(1LL * ans * x, mod);
    }
    x = MOD(1LL * x * x, mod);
    y >>= 1;
  }
  return ans;
}
int get(int l, int r, int mod) {
  if (l == r || mod == 1) {
    return MOD(w[l], mod);
  }
  return power(w[l], get(l + 1, r, phi(mod)), mod);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> w[i];
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (l == r) {
      cout << w[l] % m << "\n";
      continue;
    }
    cout << get(l, r, m) % m << "\n";
  }
  return 0;
}