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

<image>

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;
void Read(long long &x, char c = getchar(), bool f = 0) {
  for (x = 0; !isdigit(c); c = getchar())
    if (c == '-') f = 1;
  for (; isdigit(c); c = getchar()) x = 10 * x + c - '0';
  if (f) x = -x;
}
const int maxn = 1e5 + 5;
long long n, m, w[maxn];
long long MOD(long long a, long long b) { return a < b ? a : a % b + b; }
long long Qpow(long long a, long long b, long long modd) {
  long long rec = 1;
  for (a = MOD(a, modd); b; b >>= 1, a = MOD(a * a, modd))
    if (b & 1) rec = MOD(rec * a, modd);
  return rec;
}
map<long long, long long> Phi;
long long phi(long long x) {
  if (Phi.count(x)) return Phi[x];
  long long i, ans = x, tmp = x, lim = (long long)sqrt(x + 0.5);
  for (i = 2; i <= lim; i++)
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  if (x > 1) ans = ans / x * (x - 1);
  return Phi[tmp] = ans;
}
long long solve(long long l, long long r, long long modd) {
  if (l == r || modd == 1) return MOD(w[l], modd);
  return Qpow(w[l], solve(l + 1, r, phi(modd)), modd);
}
int main() {
  long long i, l, r, _;
  Read(n);
  Read(m);
  for (i = 1; i <= n; i++) Read(w[i]);
  Read(_);
  while (_--) {
    Read(l);
    Read(r);
    printf("%I64d\n", solve(l, r, m) % m);
  }
  return 0;
}
