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>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long n, m, tot, p[100005], a[100005];
long long phi(long long n) {
  long long ans = n, m = sqrt(n);
  for (long long i = 2; i <= m; i++) {
    if (n % i == 0) {
      ans = ans / i * (i - 1);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) {
    ans = ans / n * (n - 1);
  }
  return ans;
}
long long power(long long x, long long y, long long P, bool &flg) {
  if (y == 0) {
    if (1 >= P) {
      flg = true;
      return 0;
    } else
      return 1;
  }
  long long tmp = power(x, y / 2, P, flg);
  if (1ll * tmp * tmp >= P) flg = true;
  tmp = 1ll * tmp * tmp % P;
  if (y % 2 == 0)
    return tmp;
  else {
    if (1ll * tmp * x >= P) flg = true;
    return 1ll * tmp * x % P;
  }
}
void init() {
  n = read();
  p[tot = 1] = read();
  while (p[tot] != 1) {
    p[tot + 1] = phi(p[tot]);
    tot++;
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = read();
  }
  m = read();
  return;
}
void solve() {
  for (long long i = 1; i <= m; i++) {
    long long l = read(), r = read();
    long long len = r - l + 1;
    len = min(len, tot);
    long long ans = 1;
    while (len--) {
      bool flg = 0;
      ans = power(a[l + len], ans, p[len + 1], flg);
      if (flg) {
        ans += p[len + 1];
      }
    }
    printf("%I64d\n", ans % p[1]);
  }
}
int main() {
  init();
  solve();
  return 0;
}
