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.
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;


public class Main {
    static int power(long x,long n,long k){//快速求幂 x^n%k
        long res=1;
        while (n>0){
            if (n%2==0){
                x=(x*x)%k;
                n/=2;
            }else{
                n--;
                res=(res*x)%k;
            }
        }
        return (int)(res%k);
    }
    public static void main(String arg0[]){
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        long x=in.nextLong();
        long a[]=new long[n];
        long sum=0;
        for (int i=0;i<n;i++){
            a[i]=in.nextLong();
            sum+=a[i];
        }
        PriorityQueue<Long> hh=new PriorityQueue<Long>();
        for (int i=0;i<n;i++){
            hh.add(sum-a[i]);
        }
        long max=0,ge=0;
        while(!hh.isEmpty()){
            long s=hh.poll();
            max=max>s?max:s;
            ge=1;
            boolean tell=true;
            while (ge!=x){
                if (!hh.isEmpty()){
                    long g=hh.poll();
                    if (g!=s) {tell=false;break;}
                    ge++;
                }else{
                    tell=false;break;
                }
            }
            if (tell) hh.add(s+1); else break;
        }
        if (max>sum) max=sum;
        System.out.println(power(x, max, 1000000007));
        
    }
}
