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

<image>

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 INF = 0x3f3f3f3f3f3f3f3fLL;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
const double eps = 1e-6;
inline long long rd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
map<long long, long long> mp;
long long Mod(long long x, long long p) { return x <= p ? x : x % p + p; }
long long euler(long long x) {
  if (mp.find(x) != mp.end()) return mp[x];
  long long y = x;
  long long ans = x;
  long long m = sqrt(x);
  for (long long i = 2; i <= m; ++i) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans = ans / x * (x - 1);
  return mp[y] = ans;
}
long long qpow(long long a, long long b, long long p) {
  long long c = 1;
  while (b) {
    if (b & 1) c = Mod(c * a, p);
    a = Mod(a * a, p);
    b >>= 1;
  }
  return c;
}
long long w[maxn];
long long f(long long l, long long r, long long p) {
  if (l == r || p == 1) return Mod(w[l], p);
  return qpow(w[l], f(l + 1, r, euler(p)), p);
}
int main() {
  long long n = rd();
  long long m = rd();
  for (int i = 1; i <= n; ++i) {
    w[i] = rd();
  }
  int q = rd();
  while (q--) {
    long long l = rd(), r = rd();
    printf("%lld\n", f(l, r, m) % m);
  }
}