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 inf = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e6 + 4;
const int N = 5e3 + 5;
const double eps = 1e-6;
const double pi = acos(-1.0);
long long MOD(long long x, long long mod) {
  return x < mod ? x : x % mod + mod;
}
long long qpow(long long x, long long y, long long mod) {
  long long ans = 1;
  x = MOD(x, mod);
  while (y) {
    if (y & 1) ans = MOD(ans * x, mod);
    x = MOD(x * x, mod);
    y >>= 1;
  }
  return ans;
}
map<long long, long long> mp;
int n, m, w[maxn], q, l, r;
long long phi(long long m) {
  if (mp.count(m)) {
    return mp[m];
  }
  long long res = m, mm = m;
  for (long long i = 2; i * i <= m; i++) {
    if (m % i == 0) {
      res -= res / i;
      while (m % i == 0) m /= i;
    }
  }
  if (m > 1) res -= res / m;
  return mp[mm] = res;
}
long long sol(long long l, long long r, long long m) {
  if (l == r || m == 1) return MOD(w[l], m);
  return qpow(w[l], sol(l + 1, r, phi(m)), m);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    printf("%lld\n", sol(l, r, m) % m);
  }
}