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;
const int N = 100005;
int n, mod, q, l, r;
long long a[N];
map<long long, long long> M;
int phi(int n) {
  if (M.count(n) == 1) return M[n];
  int r = n, nn = n;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) {
      r = r / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  if (n > 1) r = r / n * (n - 1);
  M[nn] = r;
  return r;
}
long long qpow(long long a, long long b, long long mod = 1e18) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long check(int l, int r, int mod) {
  int pos;
  if (a[l] == 1) return 1;
  for (int i = (l); i < (min(l + 4, r) + 1); i++) {
    if (a[i] == 1) break;
    pos = i;
  }
  long long res = a[pos], x;
  for (int i = pos - 1; i >= l; i--) {
    x = 1;
    for (int j = (0); j < (res); j++) {
      x = x * a[i];
      if (x > mod) return -1;
    }
    res = x;
  }
  return res;
}
long long work(int l, int r, int mod) {
  if (mod == 1) return 0;
  if (a[l] == 1) return 1 % mod;
  if (l == r) return a[l] % mod;
  long long x = check(l + 1, r, mod);
  if (x == -1) {
    int Phi = phi(mod);
    return qpow(a[l], work(l + 1, r, Phi) + Phi, mod);
  }
  return qpow(a[l], x, mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> mod;
  for (int i = (1); i < (n + 1); i++) cin >> a[i];
  cin >> q;
  for (int cas = (0); cas < (q); cas++) {
    cin >> l >> r;
    cout << work(l, r, mod) << "\n";
  }
  return 0;
}