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 INF = 0x3f3f3f3f;
const double eps = 1e-6;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 1e6 + 10;
int n, m;
int phi[maxn];
int a[maxn];
int get(int x) {
  int res = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
int qpow(int b, int n, int mod) {
  bool flag = 0;
  int res = 1;
  while (n) {
    if (n & 1) {
      if (1LL * res * b >= mod) flag = 1;
      res = 1LL * res * b % mod;
    }
    if (1LL * b * b >= mod) flag = 1;
    b = 1LL * b * b % mod;
    n >>= 1;
  }
  if (flag) return res + mod;
  return res;
}
int solve(int l, int r, int k) {
  if (phi[k] == 1) return 1;
  if (l == r) {
    if (a[l] < phi[k])
      return a[l];
    else
      return a[l] % phi[k] + phi[k];
  } else
    return qpow(a[l], solve(l + 1, r, k + 1), phi[k]);
}
int main() {
  scanf("%d", &n);
  scanf("%d", &phi[0]);
  while (phi[m] != 1) {
    phi[m + 1] = get(phi[m]);
    m++;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    int ans = solve(l, r, 0) % phi[0];
    printf("%d\n", ans);
  }
  return 0;
}