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;
long long T, n, i = 1, x, y, a[1111111], phi[1111111];
inline long long c(long long x, long long p) { return x > p ? x % p + p : x; }
inline long long qpow(long long x, long long y, long long p) {
  long long s = 1;
  for (; y; y >>= 1, x = c(x * x, p))
    if (y & 1) s = c(s * x, p);
  return s;
}
inline long long calc(long long x) {
  long long s = x, i = 2;
  for (; i * i <= x; i++)
    if (!(x % i)) {
      s = s / i * (i - 1);
      for (; !(x % i); x /= i)
        ;
    }
  return x != 1 ? s / x * (x - 1) : s;
}
inline long long dfs(long long x, long long y, long long p) {
  return (x == y || phi[p] == 1) ? c(a[x], phi[p])
                                 : qpow(a[x], dfs(x + 1, y, p + 1), phi[p]);
}
int main() {
  for (scanf("%lld%lld", &n, phi); i <= n; i++) scanf("%lld", a + i);
  for (i = 1; i < 31; i++) phi[i] = calc(phi[i - 1]);
  for (scanf("%lld", &T); T--; printf("%lld\n", dfs(x, y, 0) % phi[0]))
    scanf("%lld%lld", &x, &y);
}