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 maxn = 1e5 + 7;
map<int, int> euler;
long long w[maxn];
int phi(int n) {
  int now = n;
  int ret = n;
  if (euler.count(now)) return euler[now];
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      ret = ret / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ret = ret / n * (n - 1);
  euler[now] = ret;
  return ret;
}
long long MOD(long long n, int mod) { return n < mod ? n : (n % mod + mod); }
long long quick_mod(long long base, long long p, int mod) {
  long long ret = 1;
  do {
    if (p & 1) ret = MOD(base * ret, mod);
    base = MOD(base * base, mod);
  } while (p >>= 1);
  return ret;
}
long long solve(int l, int r, int mod) {
  if (l == r || mod == 1) return MOD(w[l], mod);
  return quick_mod(w[l], solve(l + 1, r, phi(mod)), mod);
}
int main() {
  int n, mod;
  scanf("%d%d", &n, &mod);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long ans = solve(l, r, mod) % mod;
    printf("%lld\n", ans);
  }
}