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;
template <class T>
inline void upd1(T& a, T b) {
  a > b ? a = b : 0;
}
template <class T>
inline void upd2(T& a, T b) {
  a < b ? a = b : 0;
}
struct ano {
  operator long long() {
    long long x = 0, y = 0, c = getchar();
    while (c < 48) y = c == 45, c = getchar();
    while (c > 47) x = x * 10 + c - 48, c = getchar();
    return y ? -x : x;
  }
} buf;
int wop(int a, int n, int p) {
  int s = 1 % p;
  for (; n; n >>= 1) {
    if (n & 1) s = (long long)s * a % p;
    a = (long long)a * a % p;
  }
  return s;
}
int phi(int n) {
  int x = n;
  for (int j = 2; j * j <= n; ++j)
    if (n % j == 0) {
      x -= x / j;
      while (n % j == 0) n /= j;
    }
  if (n != 1) x -= x / n;
  return x;
}
const int N = 1e5 + 5;
int c[N], g[N], f[40];
int main() {
  int n = buf;
  f[0] = buf;
  for (int i = 0; f[i] != 1; ++i) f[i + 1] = phi(f[i]);
  for (int i = 1; i <= n; ++i) c[i] = buf;
  g[n + 1] = 1e9;
  for (int i = n; i >= 1; --i) g[i] = c[i] == 1 ? i : g[i + 1];
  int q = buf;
  while (q--) {
    int l = buf, r = min((int)buf, g[l]), s = 1;
    while (r >= l && pow(c[r], s) < 1e9) s = pow(c[r--], s) + .5;
    if (r < l)
      printf("%d\n", s % f[0]);
    else {
      function<int(int, int)> sol;
      sol = [&](int i, int j) {
        return f[j] == 1
                   ? 0
                   : wop(c[i], i == r ? s : sol(i + 1, j + 1) + f[j + 1], f[j]);
      };
      printf("%d\n", sol(l, 0));
    }
  }
}