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;
int n, m, q, a[200000];
map<int, int> p;
int Phi(int x) {
  if (p[x] != 0) return p[x];
  int tmp = x, ret = x;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      ret = (long long)ret * (i - 1) / i;
      while (x % i == 0) x = x / i;
    }
  if (x > 1) ret = (long long)ret * (x - 1) / x;
  return p[tmp] = ret, ret;
}
int Pow(int x, int p, int m) {
  if (x == 0) return 0;
  int ret = 1;
  while (p) {
    if (p & 1) ret = (long long)ret * x % m;
    x = (long long)x * x % m, p = p / 2;
  }
  return ret;
}
int cal(int x, int Y) {
  long long ret = x;
  int num = 1;
  while (ret < Y) ret *= x, num++;
  return num;
}
int ch(int l, int r, int Mo) {
  if (l > r) return 0;
  if (a[l] == 1) return Mo <= 1;
  if (a[l] >= Mo) return 1;
  return ch(l + 1, r, cal(a[l], Mo));
}
int ask(int l, int r, int Mo) {
  if (l == r || Mo == 1) return a[l];
  int P = Phi(Mo);
  int tmp = ask(l + 1, r, P);
  int pan = ch(l + 1, r, P);
  tmp = tmp + pan * P;
  return Pow(a[l], tmp, Mo);
}
int main() {
  cin >> n >> m;
  p[1] = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  cin >> q;
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << ask(l, r, m) % m << endl;
  }
}
