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;
int pi[35], phi[35], lim;
int a[100001];
void read(int &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
int get_phi(int n) {
  int ph = n;
  for (int i = 2; i * i <= n; ++i)
    if (!(n % i)) {
      ph = ph / i * (i - 1);
      while (!(n % i)) n /= i;
    }
  if (n > 1) ph = ph / n * (n - 1);
  return ph;
}
int MO(long long a, int b) { return a >= b ? a % b + b : a; }
int Pow(int a, int b, int mod) {
  int res = 1;
  while (b) {
    if (b & 1) res = MO(1LL * res * a, mod);
    a = MO(1LL * a * a, mod);
    b >>= 1;
  }
  return res;
}
int dfs(int l, int r, int dep) {
  if (dep == lim || l == r) return MO(a[l], pi[dep]);
  return Pow(a[l], dfs(l + 1, r, dep + 1), pi[dep]);
}
int main() {
  int n, mod;
  read(n);
  read(mod);
  for (int i = 1; i <= n; ++i) read(a[i]);
  int tmp = mod;
  while (pi[lim] != 1) {
    pi[++lim] = tmp;
    phi[lim] = get_phi(tmp);
    tmp = phi[lim];
  }
  int q;
  read(q);
  int l, r;
  while (q--) {
    read(l);
    read(r);
    printf("%d\n", dfs(l, r, 1) % mod);
  }
  return 0;
}