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;
long long arr[100005];
map<long long, long long> phi;
long long count_phi(long long x) {
  long long ret = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ret = ret / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ret = ret / x * (x - 1);
  return ret;
}
long long qpow(long long a, long long b, long long MOD) {
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret = ret * a;
      if (ret >= MOD) ret = ret % MOD + MOD;
    }
    b >>= 1;
    a = a * a;
    if (a >= MOD) a = a % MOD + MOD;
  }
  return ret;
}
long long dfs(long long l, long long r, long long MOD) {
  if (l == r + 1 || MOD == 1) return 1;
  return qpow(arr[l],
              dfs(l + 1, r, phi[MOD] ? phi[MOD] : phi[MOD] = count_phi(MOD)),
              MOD);
}
int main() {
  long long n, p, m;
  scanf("%lld %lld", &n, &p);
  phi[p] = count_phi(p);
  long long l, r;
  for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  scanf("%lld", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", dfs(l, r, p) % p);
  }
  return 0;
}