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 int mod = 1e9 + 7;
const double PI = 4 * atan(1.0);
const int maxm = 1e8 + 5;
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const long long LINF = 1ll << 62;
map<long long, long long> mp;
long long a[maxn];
inline long long phi(long long x) {
  long long ans = x, tmp = x;
  if (mp.count(x)) return mp[x];
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans = ans / x * (x - 1);
  mp[tmp] = ans;
  return ans;
}
inline long long qpow(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (1 & b) res = res * a < m ? res * a : (res * a % m + m);
    a = a * a < m ? a * a : (a * a % m + m);
    b >>= 1;
  }
  return res;
}
long long solve(long long index, long long r, long long m) {
  if (index == r || m == 1)
    return a[index] < m ? a[index] : (a[index] % m + m);
  else {
    long long euler = phi(m);
    long long k = solve(index + 1, r, euler);
    long long res = qpow(a[index], k, m);
    return res;
  }
}
int main() {
  long long n, m, q;
  scanf("%I64d %I64d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  scanf("%I64d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%I64d\n", solve(l, r, m) % m);
  }
  return 0;
}