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>
bool isdigit(char c) { return c >= '0' && c <= '9'; }
int read() {
  int s = 0, t = 1;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') t = -1;
  do {
    s = s * 10 + c - '0';
  } while (isdigit(c = getchar()));
  return s * t;
}
using namespace std;
const int maxn = 100010;
map<int, int> p;
int a[maxn], n, m, q;
int phi(int n) {
  if (p.count(n)) return p[n];
  int ans = n, m = n;
  for (int i = 2; i * i <= n; i++)
    if (n != 1 && n % i == 0) {
      ans = ans / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  if (n > 1) ans = ans / n * (n - 1);
  p[m] = ans;
  return ans;
}
int mod(long long x, int y) { return x < y ? x : x % y + y; }
int power(int x, int k, int m) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = mod(1ll * ans * x, m);
    x = mod(1ll * x * x, m);
    k >>= 1;
  }
  return ans;
}
int calc(int l, int r, int m) {
  if (l == r || m == 1) return mod(a[l], m);
  return power(a[l], calc(l + 1, r, phi(m)), m);
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  q = read();
  while (q--) {
    int l = read(), r = read();
    printf("%d\n", calc(l, r, m) % m);
  }
  return 0;
}
