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;
template <typename T>
bool GetMin(T &a, T b) {
  return ((a <= b) ? false : (a = b, true));
}
template <typename T>
bool GetMax(T &a, T b) {
  return ((a >= b) ? false : (a = b, true));
}
const int N = 1e5 + 10;
int n, m, q;
int w[N];
int Phi(int x) {
  static unordered_map<int, int> mp;
  if (mp.count(x)) return mp[x];
  int &res = mp[x];
  res = x;
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      res = res / i * (i - 1);
      do x /= i;
      while (x % i == 0);
    }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
int Qpow(int x, int y, int m) {
  int res = 1, fl = (pow(x, y) > m - 0.5);
  for (; y; y >>= 1, x = (long long)x * x % m)
    if (y & 1) res = (long long)res * x % m;
  return res + fl * m;
}
int Query(int l, int r, int m) {
  if (l == r) return w[l];
  if (m == 1) return 1;
  return Qpow(w[l], Query(l + 1, r, Phi(m)), m);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", w + i);
  scanf("%d", &q);
  for (int i = 1, l, r; i <= q; ++i) {
    scanf("%d%d", &l, &r);
    printf("%d\n", Query(l, r, m) % m);
  }
  return 0;
}