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 a[100005], n, mod[105], l, r, q, tot = 0;
inline long long MO(long long x, long long y) { return x > y ? x % y + y : x; }
inline long long ksm(long long x, long long y, const long long ha) {
  long long an = 1;
  for (; y; y >>= 1, x = MO(x * x, ha))
    if (y & 1) an = MO(an * x, ha);
  return an;
}
inline long long phi(long long x) {
  int tp = sqrt(x + 0.5), y = 1;
  for (int i = 2; i <= tp; i++)
    if (!(x % i)) {
      x /= i, y *= i - 1;
      while (!(x % i)) x /= i, y *= i;
      if (x == 1) break;
    }
  if (x != 1) y *= x - 1;
  return y;
}
long long solve(int now, long long mo) {
  if (now == r || mo == 1)
    return MO(a[now], mo);
  else
    return ksm(a[now], solve(now + 1, mod[now - l + 1]), mo);
}
int main() {
  scanf("%lld%lld", &n, mod);
  while (mod[tot] != 1) mod[tot + 1] = phi(mod[tot]), tot++;
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld%lld", &l, &r);
    long long ans = solve(l, mod[0]);
    if (ans >= mod[0]) ans -= mod[0];
    printf("%lld\n", ans);
  }
  return 0;
}