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;
const int maxn = 1e5 + 10;
int n, m, q;
long long mulmod(long long a, long long b, long long p) {
  if (a * b >= p)
    return a * b % p + p;
  else
    return a * b;
}
long long qpow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mulmod(res, a, p);
    b >>= 1;
    a = mulmod(a, a, p);
  }
  return res;
}
long long phi(long long n) {
  long long res = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      res /= i;
      res *= (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) {
    res /= n;
    res *= (n - 1);
  }
  return res;
}
int cnt = 0;
long long p[maxn];
long long a[maxn];
long long solve(int l, int r) {
  long long res = 1;
  for (int i = min(cnt, r - l); ~i; --i) {
    res = qpow(a[i + l], res, p[i]);
  }
  return res % p[0];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  p[cnt] = m;
  while (p[cnt] > 1) {
    p[cnt + 1] = phi(p[cnt]);
    cnt++;
  }
  p[cnt + 1] = 1;
  cin >> q;
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", solve(l, r));
  }
  return 0;
}
