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;
template <typename T>
inline T read1() {
  T t = 0;
  bool ty = 0;
  char k;
  do k = getchar(), (k == '-') && (ty = 1);
  while ('0' > k || k > '9');
  do t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  while ('0' <= k && k <= '9');
  return ty ? -t : t;
}
long long El(long long n) {
  long long tn = n;
  for (long long i = 2; i * i <= n; ++i)
    if (!(n % i)) {
      tn = tn / i * (i - 1);
      while (!(n % i)) n /= i;
    }
  if (n ^ 1) tn = tn / n * (n - 1);
  return tn;
}
long long Ge(long long n, long long m) { return n > m ? n % m + m : n; }
template <typename T>
T qkpow(T k, const long long n, long long Mod) {
  if (n == 1) return k;
  T tem = qkpow(k, n >> 1, Mod);
  if (Mod != 0) {
    tem = Ge(tem * tem, Mod);
    if (n & 1) tem = Ge(tem * k, Mod);
  } else {
    tem *= tem;
    if (n & 1) tem *= k;
  }
  return tem;
}
long long s, m, a[100001], phi[100001], l, r, tot;
long long que(long long n, long long mo) {
  if (n == r || mo == 1) return Ge(a[n], mo);
  return qkpow(a[n], que(n + 1, phi[n - l + 1]), mo);
}
signed main() {
  s = read1<long long>();
  phi[0] = read1<long long>();
  while (phi[tot] != 1) phi[tot + 1] = El(phi[tot]), ++tot;
  for (long long i = 0; i++ ^ s;) a[i] = read1<long long>();
  for (long long T = read1<long long>(); T--;) {
    l = read1<long long>(), r = read1<long long>();
    printf("%lld\n", que(l, phi[0]) % phi[0]);
  }
  return 0;
}