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>
inline int rint() {
  int x = 0;
  char s = getchar();
  for (; s < '0' || '9' < s; s = getchar())
    ;
  for (; '0' <= s && s <= '9'; s = getchar()) x = x * 10 + (s ^ '0');
  return x;
}
inline void wint(const int x) {
  if (9 < x) wint(x / 10);
  putchar(x % 10 ^ '0');
}
const int MAXN = 1e5;
int n, P, pcnt, phis[100], a[MAXN + 5];
inline int phi(int n) {
  int ret = n;
  for (int i = 2; i * i <= n; ++i) {
    if (!(n % i)) {
      ret = ret / i * (i - 1);
      for (; !(n % i); n /= i)
        ;
    }
  }
  if (n > 1) ret = ret / n * (n - 1);
  return ret;
}
inline std::pair<int, bool> mpow(const int a, const std::pair<int, bool> b,
                                 const int pid) {
  std::pair<int, bool> base(a % phis[pid], a >= phis[pid]), ret(1, false);
  long long tmp;
  for (int t = b.first + b.second * phis[pid + 1]; t; t >>= 1) {
    if (t & 1) {
      tmp = 1ll * ret.first * base.first;
      ret.first = tmp % phis[pid],
      ret.second |= base.second || tmp >= phis[pid];
    }
    tmp = 1ll * base.first * base.first;
    base.first = tmp % phis[pid], base.second |= tmp >= phis[pid];
  }
  return ret;
}
inline std::pair<int, bool> solve(const int l, const int r, const int pid) {
  if (pid == pcnt) return {1, 0};
  if (l == r) return {a[l] % phis[pid], a[l] >= phis[pid]};
  std::pair<int, bool> indx(solve(l + 1, r, pid + 1));
  return mpow(a[l], indx, pid);
}
int main() {
  n = rint(), P = rint();
  for (int t = P;; ++pcnt, t = phi(t)) {
    phis[pcnt] = t;
    if (t == 1) break;
  }
  for (int i = 1; i <= n; ++i) a[i] = rint();
  for (int q = rint(), l, r; q--;) {
    l = rint(), r = rint();
    wint(solve(l, r, 0).first % P), putchar('\n');
  }
  return 0;
}