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, w[100005], phi[105], q, cnt;
bool flag = false;
long long poww(long long a, long long b, long long p) {
  long long ans = 1;
  flag = false;
  if (a > p) {
    flag = true;
    a %= p;
  }
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      if (ans > p) {
        flag = true;
        ans %= p;
      }
    }
    a *= a;
    if (a > p) {
      flag = true;
      a %= p;
    }
    b /= 2;
  }
  if (flag) ans += p;
  return ans;
}
long long solve(long long l, long long r, long long step) {
  long long mod = phi[step], num = w[l];
  if (mod == 1) return mod;
  if (l == r) return num >= mod ? num % mod + mod : num;
  long long tmpp = solve(l + 1, r, step + 1);
  return poww(num, tmpp, mod);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> w[i];
  int tmp = m;
  while (tmp > 1) {
    cnt++;
    phi[cnt] = tmp;
    int tmpm = tmp;
    for (int i = 2; i * i <= m; i++) {
      if (tmpm % i == 0) {
        phi[cnt] = phi[cnt] * (i - 1) / i;
        while (tmpm % i == 0) tmpm /= i;
      }
    }
    if (tmpm > 1) {
      phi[cnt] = phi[cnt] * (tmpm - 1) / tmpm;
    }
    tmp = phi[cnt];
    if (tmp == 1) break;
  }
  phi[0] = m;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(l, r, 0) % m << endl;
  }
  return 0;
}
