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
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7fffffff;
const int oo = 0x3f3f3f3f;
template <typename T>
inline void rd(T &x) {
  x = 0;
  int f = 1;
  char CH = getchar();
  while (!isdigit(CH)) {
    if (CH == '-') f = 0;
    CH = getchar();
  }
  while (isdigit(CH)) {
    x = ((x + (x << 2)) << 1) + (CH - 48);
    CH = getchar();
  }
  x = (f == 1) ? x : -x;
  return;
}
const int MXN = 1e5 + 5;
int n, m, q, p;
long long a[MXN];
map<long long, long long> f;
inline long long get_phi(long long x) {
  long long ans = x, z = x;
  if (f.count(x)) return f[x];
  for (register int i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      ans -= ans / i;
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans -= ans / x;
  return f[z] = ans;
}
inline long long qpow(long long a, long long b, long long mod) {
  long long ret = 1, fl = a >= mod;
  for (; b; b >>= 1) {
    if (b & 1) {
      ret = ret * a;
      if (ret >= mod) fl = 1, ret %= mod;
    }
    a = a * a;
    if (a >= mod) a %= mod, fl = 1;
  }
  return ret + fl * mod;
}
inline long long solve(int l, int r, long long mod) {
  const long long Mod = mod;
  if (l == r) return a[l];
  if (Mod == 1) return 1;
  return qpow(a[l], solve(l + 1, r, get_phi(mod)), mod);
}
int main() {
  rd(n), rd(p);
  for (register int i = (1), i_end = (n); i <= i_end; ++i) rd(a[i]);
  for (rd(q); q--;) {
    int l, r;
    rd(l), rd(r);
    printf("%lld\n", solve(l, r, p) % p);
  }
  return 0;
}
