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 long long mod = 1e9 + 7;
;
const int N = 1e5 + 10;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const int M = 50000 + 5;
long long h[N];
map<long long, long long> mp;
long long tao(long long n) {
  long long ans = n;
  long long tt = n;
  if (mp[n]) return mp[n];
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans -= ans / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans -= ans / n;
  return mp[tt] = ans;
}
long long MOD(long long a, long long mod) {
  if (a < mod) return a;
  return a % mod + mod;
}
long long power(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = MOD(ans * a, c);
    }
    b >>= 1;
    a = MOD(a * a, c);
  }
  return ans;
}
long long solve(long long a, long long b, long long mod) {
  if (b == a) return MOD(h[a], mod);
  if (mod == 1) return MOD(h[a], mod);
  return power(h[a], solve(a + 1, b, tao(mod)), mod);
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &h[i]);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", solve(l, r, m) % m);
  }
  return 0;
}