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;
int n;
long long m;
long long a[100005];
int q;
map<long long, long long> mp;
long long phi(long long x) {
  long long xx = x;
  if (mp[x] != 0) return mp[x];
  long long ans = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) ans = ans / x * (x - 1);
  return mp[xx] = ans;
}
long long MOD(long long a, long long b) {
  if (a < b)
    return a;
  else
    return (a % b) + b;
}
long long powd(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = MOD(ans * a, mod);
    a = MOD(a * a, mod);
    b /= 2;
  }
  return ans;
}
long long solve(int l, int r, long long mod) {
  if (l == r || mod == 1) {
    return MOD(a[l], mod);
  } else {
    return powd(a[l], solve(l + 1, r, phi(mod)), mod);
  }
}
int main() {
  while (~scanf("%d%lld", &n, &m)) {
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    scanf("%d", &q);
    while (q--) {
      int l, r;
      scanf("%d%d", &l, &r);
      long long ans = solve(l, r, m);
      printf("%lld\n", ans % m);
    }
  }
  return 0;
}
