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>
const int N = 100050;
using namespace std;
inline int gi() {
  char ch = getchar();
  int x = 0, q = 0;
  while (ch < '0' || ch > '9') ch == '-' ? q = 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int f[N], g[105];
int fap(int a, int b, int mo) {
  int s = 1;
  while (b) {
    if (b & 1) {
      if (1ll * s * a >= mo)
        s = 1ll * s * a % mo + mo;
      else
        s = s * a;
    }
    b >>= 1;
    if (1ll * a * a >= mo)
      a = 1ll * a * a % mo + mo;
    else
      a = a * a;
  }
  return s;
}
int F(int n) {
  int phi = n;
  for (int i = 2; i <= n; ++i)
    if (!(n % i)) {
      phi = phi / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  return phi;
}
void work() {
  int n = gi(), q;
  int l, r, ans, m;
  g[0] = gi();
  for (int i = 1; i <= n; ++i) f[i] = gi();
  q = gi();
  for (m = 0; g[m] != 1; ++m) {
    g[m + 1] = F(g[m]);
    if (g[m] == 1) break;
  }
  for (int t = 1; t <= q; ++t) {
    l = gi(), r = gi(), ans = 1;
    for (int i = min(r, l + m); i >= l; i--) ans = fap(f[i], ans, g[i - l]);
    printf("%d\n", ans % g[0]);
  }
  return;
}
int main() {
  work();
  return 0;
}
