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
#include <bits/stdc++.h>
using namespace std;
long long n, m, a[100010], p[100010];
long long phi(long long x) {
  long long ans = 1;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans *= i - 1;
      x /= i;
      while (x % i == 0) {
        ans *= i;
        x /= i;
      }
    }
  }
  if (x > 1) ans *= x - 1;
  return ans;
}
long long Pow(long long x, long long y, long long MOD) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = (ans * x) % MOD;
    }
    x = (x * x) % MOD;
    y /= 2;
  }
  return ans;
}
long long solve(long long l, long long r, long long x) {
  long long MOD = p[x];
  if (l == r) return min(a[l] % MOD + MOD, a[l]);
  if (MOD == 1 || a[l] == 1) return 1;
  long long Po = solve(l + 1, r, x + 1);
  long long now = 1;
  for (long long i = 1; i <= Po; i++) {
    now *= a[l];
    if (now > MOD) break;
  }
  if (now <= MOD) return now;
  return Pow(a[l], Po, MOD) + MOD;
}
int main() {
  std::cin.sync_with_stdio(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  p[1] = m;
  for (long long i = 2; i <= n; i++) {
    p[i] = phi(p[i - 1]);
  }
  long long q, l, r;
  cin >> q;
  for (long long i = 1; i <= q; i++) {
    cin >> l >> r;
    cout << solve(l, r, 1) % p[1] << endl;
  }
  return 0;
}
