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;
inline long long MOD(long long a, long long mod) {
  return min(a, a % mod + mod);
}
long long quick(long long a, long long b, long long mod) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = MOD(ret * a, mod);
    a = MOD(a * a, mod);
    b >>= 1;
  }
  return MOD(ret, mod);
}
int phi(int x) {
  int ans = x;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans = ans / x * (x - 1);
  return ans;
}
vector<int> all;
vector<int> a;
long long solve(int l, int r, int m) {
  if (l == r || all[m] == 1) return MOD(a[l], all[m]);
  return quick(a[l], solve(l + 1, r, m + 1), all[m]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  int x = m;
  if (m != 1) all.push_back(m);
  while (x != 1) {
    x = phi(x);
    all.push_back(x);
  }
  all.push_back(x);
  a = vector<int>(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << solve(l, r, 0) % m << '\n';
  }
}