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

<image>

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 n, q;
long long m, w[100000];
vector<long long> phiChain;
long long modPow(long long a, long long p, long long mod) {
  return (p ? modPow((a * a) % mod, p / 2, mod) * ((p & 1) ? a : 1) : 1) % mod;
}
long long phi(long long x) {
  long long temp = x;
  for (int i = 2; i * i <= temp; i++) {
    if (temp % i == 0) x = x / i * (i - 1);
    while (temp % i == 0) temp /= i;
  }
  if (temp > 1) x = x / temp * (temp - 1);
  return x;
}
long long getRemaining(int l, int r) {
  if (w[l] == 1) return 1;
  if (w[l] >= 50) return 50;
  if (r == l + 1) return w[l];
  int z = 1, c = 0;
  while (z <= 50) z *= w[l], c++;
  int re = getRemaining(l + 1, r);
  if (re >= c)
    return 50;
  else
    return modPow(w[l], re, 100);
}
long long solve(int l, int r, int idx) {
  if (r == l + 1) return w[l] % phiChain[idx];
  if (phiChain[idx] == 1) return 0;
  int remaining = getRemaining(l + 1, r);
  if (remaining < 50) return modPow(w[l], remaining, phiChain[idx]);
  return modPow(w[l], phiChain[idx + 1] + solve(l + 1, r, idx + 1),
                phiChain[idx]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < (n); i++) cin >> w[i];
  phiChain.push_back(m);
  while (phiChain.back() > 1) phiChain.push_back(phi(phiChain.back()));
  cin >> q;
  for (int i = 0; i < (q); i++) {
    int l, r;
    cin >> l >> r;
    l--;
    cout << solve(l, r, 0) << "\n";
  }
  return 0;
}