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;
long long qr() {
  long long sum = 0, b = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') b = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = (sum << 1) + (sum << 3) + ch - '0';
    ch = getchar();
  }
  return sum * b;
}
long long n, t, m, a[100010];
map<long long, long long> ph;
long long phi(long long n) {
  long long ans = n, m = sqrt(n);
  for (long long i = 2; i <= m; ++i) {
    if (n % i == 0) {
      ans = (ans / i) * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans = (ans * (n - 1)) / n;
  return ans;
}
long long qkpow(long long a, long long b, long long mo) {
  long long an = 1;
  while (b) {
    if (b & 1) {
      an = an * a;
      if (an >= mo) an = an % mo + mo;
    }
    a = a * a;
    if (a >= mo) a = a % mo + mo;
    b >>= 1;
  }
  return an;
}
long long l, r;
long long ddfs(long long i, long long mo) {
  if (i == r + 1 || mo == 1) return 1;
  long long t = ddfs(i + 1, ph[mo]);
  return qkpow(a[i], t, mo);
}
signed main() {
  n = qr(), m = qr();
  for (long long i = 1; i <= n; ++i) a[i] = qr();
  t = qr();
  long long cu = m;
  while (cu != 1) ph[cu] = phi(cu), cu = ph[cu];
  ph[cu] = 1;
  for (long long o = 1; o <= t; ++o) {
    l = qr(), r = qr();
    printf("%lld\n", ddfs(l, m) % m);
  }
  return 0;
}