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.io.*;
import java.math.*;
import static java.lang.Math.*;
import java.util.*;


public class Main {

	static long oo = 2000000000000000000L;

	public static void main(String[] args)throws IOException {
		BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(buf.readLine()); 

		int n = Integer.parseInt(st.nextToken());
		int p = Integer.parseInt(st.nextToken());

		st = new StringTokenizer(buf.readLine());

		long sum = 0;

		long []a = new long[n + 1];

		for(int i = 0; i < n; i++){
			a[i] = Long.parseLong(st.nextToken());
			sum += a[i];
		}

		for(int i = 0; i < n; i++) a[i] = sum - a[i];

		a[n] = oo;

		Arrays.sort(a);

		long pot = a[0];
		long res = 1;
		long act = a[0];

		for(int i = 1;  i <= n; i++){
			if( act == a[i] )res++;
			else{
				act = a[i];
				long cant = 0;
				long tmp = res;
				while( tmp % p == 0 ){
					cant ++; tmp /= p;
				} 
				if(cant == 0)break;

				if(cant < act - pot){
					pot += cant; break; 
				}else{
					res /= (long)(pow(p, act - pot));
					pot += (act - pot);
					res ++;
				}
			}
		}

		pot = min(pot,sum);

		BigInteger aa = BigInteger.valueOf(p).modPow(BigInteger.valueOf(pot), BigInteger.valueOf(1000000007));
		System.out.println(aa);
	}
}
