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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, q, P, a[100010];
int phi(int n) {
  int ans = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans = ans / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  return n != 1 ? ans / n * (n - 1) : ans;
}
vector<int> ps;
int exp(int a, int n, int P) {
  int res = 1;
  for (; n; n >>= 1) {
    if (n & 1)
      res =
          min((unsigned long long)res * a % P + P, (unsigned long long)res * a);
    a = min((unsigned long long)a * a % P + P, (unsigned long long)a * a);
  }
  return res;
}
int query(int l, int r, int pn) {
  if (l > r) return 1;
  if (pn == ps.size() - 1) return min(a[r], 2);
  return exp(a[l], query(l + 1, r, pn + 1), ps[pn]);
}
int main() {
  int i;
  n = read();
  P = read();
  for (i = 1; i <= n; i++) a[i] = read();
  while (P > 1) ps.push_back(P), P = phi(P);
  ps.push_back(1);
  q = read();
  while (q--) {
    int l = read(), r = read();
    printf("%d\n", query(l, r, 0) % ps[0]);
  }
  return 0;
}