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.
//package com.company;

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
	    new Main().run();
    }

    public void run() throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        long  n,x;
        String [] line=in.readLine().split(" ");
        n=Long.parseLong(line[0]);
        x=Long.parseLong(line[1]);
        line=in.readLine().split(" ");
        ArrayList <Long> a = new ArrayList(),b=new ArrayList();
        long sum=0;
        for (int i = 0; i < n;i++){
            a.add(Long.parseLong(line[i]));
            sum+=a.get(i);
        }
        for (int i = 0; i < n;i++){
            b.add(sum - a.get(i));
        }
        Collections.sort(b);
        Collections.reverse(b);
        while (true){
            long v=b.get(b.size()-1);
            long cnt=0;
            while (b.size() > 0 && b.get(b.size() - 1) == v) {
                cnt ++;
                b.remove(b.size()-1);
            }

            if (cnt % x != 0) {
                v = Math.min(v, sum);
                System.out.println(bpow(x, v));
                break;
            }
            else {
                cnt /= x;
                for(int i = 0; i < cnt; i++)
                    b.add(v + 1);
            }
        }

    }
    long bpow(long x,long v){
        if (v == 0)
            return 1 % 1000000007;
        if (v % 2 == 1)
            return (x * bpow(x, v - 1)) % 1000000007;

        long r = bpow(x, v / 2);
        return (r * r) % 1000000007;
    }

}
