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;
const long long MAXN = 100005;
long long read() {
  long long x = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x * flag;
}
long long n, m, q, l, r, a[MAXN];
map<long long, long long> ph;
long long phi(long long n) {
  long long t = sqrt(n), res = n;
  for (long long i = 2; i <= t; i++) {
    if (n % i == 0) {
      res /= i;
      res *= (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) res /= n, res *= (n - 1);
  return res;
}
long long qkpow(long long a, long long b, long long MOD) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
      if (res >= MOD) res = res % MOD + MOD;
    }
    a = a * a;
    if (a >= MOD) a = a % MOD + MOD;
    b >>= 1;
  }
  return res;
}
long long dfs(long long x, long long p) {
  if (x == r + 1 || p == 1) return 1;
  long long t = dfs(x + 1, ph[p]);
  return qkpow(a[x], t, p);
}
signed main() {
  n = read();
  m = read();
  long long cur = m;
  while (cur != 1) ph[cur] = phi(cur), cur = ph[cur];
  ph[cur] = 1;
  for (long long i = 1; i <= n; i++) a[i] = read();
  q = read();
  while (q--) {
    l = read();
    r = read();
    printf("%lld\n", dfs(l, m) % m);
  }
}