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 maxn = 100010;
int n, q;
long long P, a[maxn];
long long qp(long long x, long long y, long long P) {
  long long z = 1, flag = x >= P;
  for (; y; y >>= 1) {
    if ((y & 1) && (z *= x) >= P) flag = 1, z %= P;
    if ((x *= x) >= P) flag = 1, x %= P;
  }
  return z + flag * P;
}
int main() {
  scanf("%d %lld", &n, &P);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  auto phi = [&](long long x) {
    static unordered_map<long long, long long> mp;
    if (mp.count(x)) return mp[x];
    long long y = x, z = x;
    for (long long i = 2; i * i <= x; i++)
      if (!(x % i)) {
        y = y / i * (i - 1);
        while (!(x % i)) x /= i;
      }
    if (x > 1) y = y / x * (x - 1);
    return mp[z] = y;
  };
  function<long long(int, int, long long)> solve = [&](int l, int r, int P) {
    if (l == r) return a[l];
    if (P == 1) return 1LL;
    return qp(a[l], solve(l + 1, r, phi(P)), P);
  };
  scanf("%d", &q);
  for (int i = 1, l, r; i <= q; i++) {
    scanf("%d %d", &l, &r);
    printf("%lld\n", solve(l, r, P) % P);
  }
  return 0;
}