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;
const int N = 2e5 + 5;
int mod;
int n, m, t, q;
long long a[N], eul[100];
int euler(int n) {
  int res = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) res = res / i * (i - 1);
    while (n % i == 0) n /= i;
  }
  if (n > 1) res = res / n * (n - 1);
  return res;
}
long long jiangmi(long long x, long long mod) {
  return x >= mod ? x % mod + mod : x;
}
long long quickpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = jiangmi(ans * a, mod);
    a = jiangmi(a * a, mod);
    b >>= 1;
  }
  return jiangmi(ans, mod);
}
long long dfs(int l, int r, int i) {
  if (l == r || eul[i] == 1) return jiangmi(a[l], eul[i]);
  return quickpow(a[l], dfs(l + 1, r, i + 1), eul[i]);
}
int main() {
  scanf("%d%d", &n, &mod);
  eul[0] = mod;
  for (int i = (1); i <= (30); i++) eul[i] = euler(eul[i - 1]);
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  scanf("%d", &m);
  int l, r;
  for (int i = (1); i <= (m); i++) {
    scanf("%d%d", &l, &r);
    printf("%lld\n", dfs(l, r, 0) % mod);
  }
  return 0;
}