Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number <image> on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: <image>, where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109). 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
2 2


Output

8


Input

3 3
1 2 3


Output

27


Input

2 2
29 29


Output

73741817


Input

4 5
0 0 0 0


Output

1

Note

In the first sample <image>. Thus, the answer to the problem is 8.

In the second sample, <image>. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample <image>. Thus, the answer to the problem is 1.
#include <bits/stdc++.h>
using namespace std;
long long int modulo = 1000000007;
long long int eleva(long long int b, long long int e) {
  if (e == 0) return 1;
  if (e % 2 == 0) return eleva(b * b % modulo, e / 2);
  return b * eleva(b * b % modulo, e / 2) % modulo;
}
long long int a[1000000];
int main() {
  ios::sync_with_stdio(false);
  long long int n, x;
  cin >> n >> x;
  long long int s = 0;
  long long int maxim = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    s += a[i];
    maxim = max(maxim, a[i]);
  }
  map<long long int, long long int> veces;
  for (int i = 0; i < n; i++) veces[maxim - a[i]]++;
  for (;;) {
    map<long long int, long long int>::iterator it = veces.begin();
    long long int aa = it->first;
    long long int vec = it->second;
    if (vec % x != 0) break;
    long long int anyade = vec / x;
    veces.erase(it);
    veces[aa + 1] += anyade;
  }
  map<long long int, long long int>::iterator it = veces.begin();
  long long int minimo = it->first;
  long long int e = s - maxim + min(minimo, maxim);
  cout << eleva(x, e) << endl;
}
