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 <typename T>
inline void upmin(T &x, T y) {
  y < x ? x = y : 0;
}
template <typename T>
inline void upmax(T &x, T y) {
  x < y ? x = y : 0;
}
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long OO = 1e18;
const int N = 1e5 + 100;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int w[N];
int p[N];
inline int phi(int n) {
  int ans = n;
  for (int i = 2; i <= n; i++)
    if (!(n % i)) {
      ans = ans / i * (i - 1);
      while (!(n % i)) n /= i;
    }
  return ans;
}
inline int mul(int x, int y, int mod) {
  return 1LL * x * y >= mod ? 1LL * x * y % mod + mod : x * y;
}
inline int qpow(int x, int y, int mod) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x, mod);
    x = mul(x, x, mod);
    y >>= 1;
  }
  return ans;
}
int main() {
  int n = gi(), i, Q, m, l, r, ans;
  p[0] = gi();
  for (m = 0; p[m] != 1; p[m + 1] = phi(p[m]), m++)
    ;
  p[m + 1] = 1;
  for (i = 1; i <= n; i++) w[i] = gi();
  Q = gi();
  while (Q--) {
    l = gi(), r = gi();
    ans = 1;
    for (i = min(r - l, m); ~i; i--) ans = qpow(w[i + l], ans, p[i]);
    cout << ans % p[0] << endl;
  }
  return 0;
}