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;
using ll = long long;
long long okay(long long val, long long mod) {
  if (val < mod) {
    return val;
  }
  return (val % mod) + mod;
}
long long powM(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = okay(1LL * res * a, mod);
    }
    a = okay(1LL * a * a, mod);
    b >>= 1;
  }
  return res;
}
long long sz = 0;
const long long N = 1e5 + 10;
long long mod[5000];
long long arr[N];
long long phi(long long a) {
  long long ans = a;
  for (long long i = 2; i * i <= a; i++) {
    long long cnt = 0;
    while (a % i == 0) a /= i, cnt++;
    if (cnt != 0) ans -= ans / i;
  }
  if (a != 1) {
    ans -= ans / a;
  }
  return ans;
}
long long findans(long long l, long long r, long long cur) {
  if (cur == sz) return 1;
  if (l == r) return okay(arr[l], mod[cur]);
  long long x = findans(l + 1, r, cur + 1);
  long long p = x % mod[cur];
  if (x >= mod[cur]) p = p + mod[cur];
  long long ans = powM(arr[l], p, mod[cur]);
  return ans;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  mod[sz] = m;
  while (mod[sz] != 1) {
    sz++;
    mod[sz] = phi(mod[sz - 1]);
  }
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << findans(l, r, 0) % m << endl;
  }
}
