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;
inline int read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long n, p, m, A[500010];
long long phi[500010], g[500010];
inline long long ksm(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
inline pair<long long, long long> query(long long l, long long r, long long a) {
  long long v = A[l];
  if (v == 0) return make_pair(0, 0);
  if (g[a] == 1) return make_pair(0, 1);
  if (v == 1) return make_pair(1, 0);
  if (l == r) return v < g[a] ? make_pair(v, 0) : make_pair(v % g[a], 1);
  pair<long long, long long> t = query(l + 1, r, a + 1);
  if (t.second)
    return make_pair(ksm(v % g[a], t.first + (g[a + 1]), g[a]), 1);
  else {
    if (v >= g[a])
      return make_pair(ksm(v % g[a], t.first, g[a]), 1);
    else {
      long long val = 1, tag = 0;
      for (register long long i = 1; i <= t.first; i++) {
        val *= v;
        if (val >= g[a]) {
          tag = 1;
          break;
        }
      }
      if (!tag)
        return make_pair(ksm(v, t.first, g[a]), 0);
      else
        return make_pair(ksm(v, t.first, g[a]), 1);
    }
  }
}
inline long long QPHI(long long n) {
  if (n == 1) return 1;
  long long ans = n, tmp = n;
  for (register long long i = 2; i * i <= n; i++) {
    if (tmp % i == 0) {
      ans = ans / i * (i - 1);
      while (tmp % i == 0) tmp /= i;
    }
  }
  if (tmp > 1) ans = ans / tmp * (tmp - 1);
  return ans;
}
int main() {
  n = read(), p = read();
  for (register long long i = 1; i <= n; i++) A[i] = read();
  g[0] = p;
  for (register long long i = 1; i <= 100; i++) g[i] = QPHI(g[i - 1]);
  m = read();
  while (m--) {
    long long l = read(), r = read();
    printf("%lld\n", query(l, r, 0).first);
  }
  return 0;
}