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;
long long w[100010];
map<long long, long long> mp;
long long qpow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x < mod ? res * x : res * x % mod + mod, n--;
    x = x * x < mod ? x * x : x * x % mod + mod;
    n >>= 1;
  }
  return res;
}
long long phi(long long m) {
  long long s = m;
  if (mp.count(m)) return mp[m];
  for (long long i = 2; i * i <= m; i++) {
    if (m % i == 0) s = s / i * (i - 1);
    while (m % i == 0) m /= i;
  }
  if (m > 1) s = s / m * (m - 1);
  mp[m] = s;
  return s;
}
long long solve(long long l, long long r, long long m) {
  if (l == r || m == 1) {
    if (w[l] < m)
      return w[l];
    else
      return ((w[l] % m) + m);
  }
  return qpow(w[l], solve(l + 1, r, phi(m)), m);
}
int main() {
  long long n, m, q;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> w[i];
  cin >> q;
  while (q > 0) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r, m) % m << endl;
    q--;
  }
  return 0;
}