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;
const int MX = 100005;
int n, m, w[MX], q, l, r;
bitset<MX> bs;
vector<long long> pr;
void criba() {
  for (long long i = 2; i < MX; i++)
    if (!bs[i]) {
      pr.push_back(i);
      for (long long j = i * i; j < MX; j += i) bs[j] = 1;
    }
}
map<int, int> mp;
long long phi(long long n) {
  if (mp.count(n)) return mp[n];
  long long res = n, aux = n;
  int i = 0;
  while (pr[i] * pr[i] <= n) {
    if (n % pr[i] == 0) res -= res / pr[i];
    while (n % pr[i] == 0) n /= pr[i];
    i++;
  }
  if (n != 1) res -= res / n;
  return mp[aux] = res;
}
long long mod(long long x, long long y) {
  if (x < y) return x;
  return x % y + y;
}
long long pot(long long b, long long p, long long m) {
  long long res = 1;
  while (p) {
    if (p & 1) res = mod(res * b, m);
    b = mod(b * b, m);
    p /= 2;
  }
  return res;
}
long long obtRes(int l, int r, int m) {
  if (l == r || m == 1) return mod(w[l], m);
  return pot(w[l], obtRes(l + 1, r, phi(m)), m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  criba();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> w[i];
  cin >> q;
  while (q--) {
    cin >> l >> r;
    cout << obtRes(l, r, m) % m << '\n';
  }
  return 0;
}