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> w;
long long a[((long long)(2e6 + 900))];
long long n, m;
long long q;
long long tot(long long x) {
  long long p = x;
  long long c = x;
  for (long long i = 2; i * i <= c; i++) {
    if (c % i == 0) {
      p /= i;
      p *= i - 1;
      while (c % i == 0) c /= i;
    }
  }
  if (c > 1) {
    p /= c;
    p *= c - 1;
  }
  return p;
}
long long mmod(long long x, long long m) {
  if (x >= m) return m + (x) % m;
  return x;
}
long long pw(long long a, long long b, long long m) {
  long long p = 1;
  while (b) {
    if (b & 1ll) p = mmod((p * a), m);
    a = mmod((a * a), m);
    b >>= 1ll;
  }
  return p;
}
long long get(long long l, long long r, long long m) {
  if (l == r) return mmod(a[l], m);
  if (m == 1) return 1;
  long long k = get(l + 1, r, w[m]);
  return pw(a[l], k, m);
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long k = m;
  while (k > 1) {
    w[k] = tot(k);
    k = w[k];
  }
  long long q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    l--;
    r--;
    cout << get(l, r, m) % m << endl;
  }
}