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;
const int N = 100010;
long long calc(long long x) {
  long long i, tmp = x, n = x;
  for (i = 2; i * i <= n; i++) {
    if (x % i != 0) continue;
    tmp -= tmp / i;
    while (x % i == 0) x /= i;
  }
  if (x != 1) tmp -= tmp / x;
  return tmp;
}
long long ksm(long long a, long long x, long long p) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans *= a, ans = (ans >= p ? ans % p + p : ans);
    a *= a, a = (a >= p ? a % p + p : a), x >>= 1;
  }
  return ans;
}
long long l, r, phi[N], a[N], n, q, mod;
long long ff(long long x, long long p) {
  if (x == r + 1 || p == 1) return 1;
  long long t = phi[x - l + 1];
  return ksm(a[x], ff(x + 1, t), p);
}
int main() {
  long long i, j, tmp;
  scanf("%lld%lld", &n, &mod);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  tmp = mod;
  for (i = 1; tmp != 1; i++) tmp = phi[i] = calc(tmp);
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", ff(l, mod) % mod);
  }
  return ~~(0 - 0);
}