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 practice {
		public static void main(String[] args) {
		Scanner scn=new Scanner(System.in);
		int n=scn.nextInt(),x=scn.nextInt();
		int[] arr=new int[n+1];
		long sum=0;
		for(int i=1;i<=n;i++){
			arr[i]=scn.nextInt();
			sum+=arr[i];
		}
		long max=arr[n],com=1;
		for(int i=n-1;i>=1;i--){
			while(arr[i]<max&&com%x==0){
				max--;
				com/=x;
			}
			if(arr[i]==max){
				com++;
			}
			else{
				break;
			}
		}
		while(com%x==0&&max>0){
			max--;
			com/=x;
		}
		
		long sol=pow(x,sum-max);
		System.out.println(sol);
		}
		public static long pow(long a,long b){
			long t=1;
			 int mod=1000000007;
			 while(b>0){
				 if((b&1)>0){
						t=t*a%mod;
					}
				 a=a*a%mod;
				 b>>=1;
			 }
			return t;
			
		}
	}

