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
#include <bits/stdc++.h>
using namespace std;
map<long long, long long> E;
long long euler_phi(long long n) {
  if (n == 0) return 0;
  long long ans = n;
  for (long long x = 2; x * x <= n; ++x)
    if (n % x == 0) {
      ans -= ans / x;
      while (n % x == 0) n /= x;
    }
  if (n > 1) ans -= ans / n;
  return ans;
}
long long inline smod(long long x, long long d) {
  return (x >= d) ? d + x % d : x;
}
long long spow(long long x, long long p, long long mod) {
  if (p == 0) return 1;
  long long r = spow(x, p / 2, mod);
  return smod(smod(r * r, mod) * (p % 2 == 1 ? x : 1), mod);
}
vector<long long> phs;
long long mult_pow(vector<long long> &v, int l, int r, long long mod) {
  long long ans = 1;
  for (int i = min((r - l), (int)phs.size() - 1); i >= 0; --i) {
    ans = spow(v[l + i], ans, phs[i]);
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, mod;
  cin >> n >> mod;
  m = mod;
  while (m != 1 && phs.size() < 100000) phs.push_back(m), m = euler_phi(m);
  phs.push_back(1);
  vector<long long> v(n);
  for (auto &x : v) cin >> x;
  int q, a, b;
  cin >> q;
  while (q--) {
    cin >> a >> b;
    --a, --b;
    cout << mult_pow(v, a, b, mod) << '\n';
  }
}
