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 = 101000;
int n, m, q;
int que[maxn], cnt;
int num[maxn];
inline int read() {
  int res = 0;
  char ch = getchar(), ch1 = ch;
  while (!isdigit(ch)) ch1 = ch, ch = getchar();
  while (isdigit(ch)) res = (res << 3) + (res << 1) + ch - '0', ch = getchar();
  return ch1 == '-' ? -res : res;
}
int phi(int p) {
  int ans = p;
  for (register int i = 2; i * i <= p; ++i) {
    if (!(p % i)) {
      ans = ans / i * (i - 1);
      while (!(p % i)) p /= i;
    }
  }
  if (p > 1) ans = ans / p * (p - 1);
  return ans;
}
long long mod(long long x, long long p) { return x < p ? x : (x % p + p); }
int qpow(int x, int p, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = mod((long long)res * x, p);
    x = mod((long long)x * x, p);
    k >>= 1;
  }
  return res;
}
long long query(int l, int r, int st) {
  if (!que[st]) return 1;
  if (l == r) return mod(num[l], que[st]);
  long long rans = query(l + 1, r, st + 1);
  return qpow(num[l], que[st], rans);
}
int main() {
  n = read();
  m = read();
  que[0] = m;
  while (que[cnt] != 1) que[cnt + 1] = phi(que[cnt]), ++cnt;
  for (register int i = 1; i <= n; ++i) num[i] = read();
  q = read();
  for (register int i = 0, l, r; i < q; ++i) {
    l = read();
    r = read();
    printf("%lld\n", query(l, r, 0) % m);
  }
  return 0;
}