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 a[100010], t[100010], f[10010];
int n, m, q, l, r, cnt;
int phi(int x) {
  int sz = sqrt(x);
  memset(t, 0, sizeof(t));
  int tot = 0;
  int ans = x;
  for (int i = 2; i <= sz; i++) {
    if (x % i == 0) t[++tot] = i;
    while (x % i == 0) x /= i;
  }
  if (x != 1) t[++tot] = x;
  for (int i = 1; i <= tot; i++) ans = ans / t[i] * (t[i] - 1);
  return ans;
}
long long mode(long long a, long long b) {
  if (a < b) return a;
  return a % b + b;
}
int solve(int l, int r, int dep) {
  if (l > r || dep >= cnt) return 1;
  int ep = solve(l + 1, r, dep + 1);
  long long nowx = a[l];
  long long ans = 1;
  for (; ep; ep >>= 1, nowx = mode(nowx * nowx, f[dep]))
    if (ep & 1) ans = mode(ans * nowx, f[dep]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  f[++cnt] = m;
  while (f[cnt] != 1) f[cnt + 1] = phi(f[cnt]), cnt++;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    printf("%d\n", solve(l, r, 1) % m);
  }
}
