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 maxm = 1e5 + 10;
int n, m, q, bj;
int v[maxm + 10];
vector<int> M;
inline int work(int a, int b, int m) {
  if (1ll * a * b >= m) bj = 1;
  return 1ll * a * b % m;
}
int qpow(int a, int b, int m) {
  int re = 1;
  for (bj = 0; b; b >>= 1, a = work(a, a, m))
    if (b & 1) re = work(re, a, m);
  return re;
}
int getphi(int t) {
  int re = t;
  for (int i = 2; i * i <= t; i++) {
    if (t % i != 0) continue;
    re /= i, re *= (i - 1);
    while (t % i == 0) t /= i;
  }
  if (t != 1) re /= t, re *= (t - 1);
  return re;
}
int euler(int v[], int l, int r) {
  bj = 0;
  int mx = min(r - l + 1, (int)M.size());
  int now = work(v[l + mx - 1], 1, M[mx - 1]);
  for (int i = mx - 2; i >= 0; i--) {
    if (bj) now += M[i + 1];
    now = qpow(v[l + i], now, M[i]);
  }
  return now;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  int tmp = m;
  for (M.push_back(tmp); tmp != 1; tmp = getphi(tmp), M.push_back(tmp))
    ;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", euler(v, l, r));
  }
  return 0;
}