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>
using namespace std;
long long n, m, p;
long long w[1000005];
long long get_phi(long long n) {
  long long res = n;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i) continue;
    while (n % i == 0) n /= i;
    res = res * (i - 1) / i;
  }
  if (n > 1) res = res / n * (n - 1);
  return res;
}
long long phi[105];
void init() {
  phi[0] = p;
  for (int i = 1; i <= 50; ++i) {
    phi[i] = get_phi(phi[i - 1]);
  }
}
long long Mod(long long x, long long mod) {
  if (x <= mod)
    return x;
  else
    return x % mod + mod;
}
long long qpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = Mod(res * a, mod);
    a = Mod(a * a, mod);
    n >>= 1;
  }
  return res;
}
long long solve(long long l, long long r, long long st) {
  if (l == r || phi[st] == 1) {
    return Mod(w[l], phi[st]);
  }
  return qpow(w[l], solve(l + 1, r, st + 1), phi[st]);
}
int main() {
  cin >> n >> p;
  init();
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &w[i]);
  }
  cin >> m;
  int l, r;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &l, &r);
    printf("%lld\n", solve(l, r, 0) % p);
  }
  return 0;
}