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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int get() {
  register char c = getchar();
  register int o = 0;
  while (c < 48) c = getchar();
  while (c > 47) o = o * 10 + c - 48, c = getchar();
  return o;
}
int O[20], B;
void out(register long long x) {
  if (!x) {
    puts("0");
    return;
  }
  while (x) O[++B] = x % 10, x /= 10;
  while (B) putchar(O[B--] + 48);
  putchar('\n');
}
int n, m, w[100005], l[100005], r[100005], f[105], s;
long long ok;
long long fpower(register long long x, register long long k,
                 register long long p) {
  register long long ans = 1;
  ok = 0;
  while (k) {
    if (k & 1) {
      ans *= x;
      if (ans >= p) ans %= p, ok = p;
    }
    k >>= 1;
    x *= x;
    if (x >= p) x %= p, ok = p;
  }
  return ans + ok;
}
long long sol(register int L, register int R, register int K) {
  if (L == R) {
    if (w[L] < f[K]) {
      ok = 0;
      return w[L] % f[K];
    } else {
      ok = f[K];
      return w[L] % f[K] + f[K];
    }
  }
  return fpower(w[L], sol(L + 1, R, K + 1), f[K]);
}
int main() {
  n = get(), m = get();
  for (register int i = 1; i <= n; ++i) w[i] = get();
  f[0] = m;
  while (f[s] > 1) {
    register int t = f[s], T = t;
    for (register int i = 2; i * i <= T; ++i) {
      if (T % i == 0) {
        t = t / i * (i - 1);
        while (T % i == 0) T /= i;
      }
    }
    if (T > 1) t = t / T * (T - 1);
    f[++s] = t;
  }
  f[++s] = 1, f[++s] = 1;
  register int q = get();
  for (register int i = 1; i <= q; ++i)
    l[i] = get(), r[i] = min(l[i] + s, get());
  for (register int i = 1; i <= q; ++i) ok = 0, out(sol(l[i], r[i], 0) - ok);
  return 0;
}