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

<image>

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 n, l, r;
long long p, a[100005], ph[100005];
inline long long read() {
  long long s = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  return s;
}
long long phi(long long p) {
  long long x = p, rt = p;
  for (register int i = 2; i * i <= p; i++)
    if (x % i == 0) {
      rt = rt / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  if (x > 1) rt = rt / x * (x - 1);
  return rt;
}
long long MOD(long long x, long long mod) {
  return x >= mod ? (x % mod + mod) : x;
}
long long ksm(long long a, long long k, long long p) {
  long long rt = 1;
  while (k) {
    if (k & 1) rt = MOD(rt * a, p);
    a = MOD(a * a, p);
    k >>= 1;
  }
  return rt;
}
long long solve(long long st, long long ed, long long p) {
  if (ph[p] == 1 || st == ed) return MOD(a[st], ph[p]);
  return ksm(a[st], solve(st + 1, ed, p + 1), ph[p]);
}
int main() {
  n = read();
  ph[0] = read();
  for (register int i = 1; i <= n; i++) a[i] = read();
  for (register int i = 1; i <= n; i++) ph[i] = phi(ph[i - 1]);
  int q = read();
  while (q--) {
    l = read(), r = read();
    if (l == r)
      printf("%lld\n", a[l] % ph[0]);
    else
      printf("%lld\n", solve(l, r, 0) % ph[0]);
  }
  return 0;
}