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.*;
public class PrimeNumber {
	private static final long MOD=1000000007;
	public static void main(String[]args){
		Scanner sc=new Scanner(System.in);
		long n=sc.nextLong(),x=sc.nextLong(),sum=0;
		TreeMap<Integer,Integer>pows=new TreeMap<Integer,Integer>();
		for(int i=0;i<n;i++){
			int a=sc.nextInt();
			sum+=a;
			boolean b;
			do{
				b=false;
				if(!pows.containsKey(a))
					pows.put(a,1);
				else
					pows.put(a,pows.get(a)+1);
				if(pows.get(a)==x&&a>0){
					pows.remove(a--);
					b=true;
				}
			}while(b);
		}
		sc.close();
		System.out.println(power(x,(sum-pows.lastKey())%(MOD-1)));
	}
	public static long power(long base, long pow){
		if(pow==0)
			return 1;
		if(pow==1)
			return base;
		long p=power(base,pow/2);
		return ((p*p)%MOD)*(pow%2==1?base:1)%MOD;
	}
}