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

<image>

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>
long long phi(long long a) {
  long long ans;
  ans = a;
  for (long long i = 2; i * i <= a; i++) {
    bool f(0);
    while (!(a % i) && a != 1) f = 1, a /= i;
    if (f) ans -= ans / i;
  }
  if (a != 1) ans -= ans / a;
  return ans;
}
bool pd;
long long pow(long long a, long long b, long long m) {
  pd = 0;
  long long ans(1);
  while (b) {
    if (b & 1) {
      if (ans * a >= m) pd = 1;
      ans = ans * a % m;
    }
    if (a * a >= m && b != 1) pd = 1;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long m, m1[1000000], n, a[1000000], k;
long long tow(long long l, long long r, long long i) {
  if (l == r) {
    pd = 0;
    if (a[l] >= m1[i]) pd = 1;
    return a[l] % m1[i];
  }
  if (m1[i] == 1) {
    pd = 1;
    return 0;
  }
  long long v1 = tow(l + 1, r, i + 1);
  if (pd) v1 += m1[i + 1];
  return pow(a[l], v1, m1[i]);
}
int main() {
  scanf("%lld %lld", &n, &m);
  m1[1] = m;
  for (int i = 2; i <= n; i++) {
    m1[i] = phi(m1[i - 1]);
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  scanf("%lld", &k);
  while (k--) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", tow(l, r, 1));
  }
  return 0;
}