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 N = 101010;
const long long qwq = 303030;
const long long inf = 0x3f3f3f3f;
long long n, m, mp;
long long a[N];
map<long long, long long> f;
inline long long read() {
  long long sum = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum * f;
}
long long fai(long long now) {
  if (f[now]) return f[now];
  long long sq = sqrt(now), res = now, u = now;
  for (long long i = 2; i <= sq; i++) {
    if (now % i) continue;
    res = res / i * (i - 1);
    while (now % i == 0) now /= i;
  }
  if (now != 1) res = res / now * (now - 1);
  return f[u] = res;
}
long long ksm(long long aa, long long bb, long long p) {
  long long sum = 1, chao = 0;
  while (bb) {
    if (bb & 1) sum = sum * aa;
    bb >>= 1;
    aa = aa * aa;
    if (sum >= p || aa >= p) chao = 1;
    sum %= p;
    aa %= p;
  }
  return sum + (chao ? p : 0);
}
long long cheng(long long l, long long r, long long p) {
  if (p == 1) return 1;
  if (l == r + 1) return 1;
  return ksm(a[l], cheng(l + 1, r, fai(p)), p);
}
int main() {
  long long x, y;
  n = read();
  mp = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  m = read();
  while (m--) {
    x = read();
    y = read();
    cout << cheng(x, y, mp) % mp << "\n";
  }
  return 0;
}