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.BufferedReader;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class PrimeNumber {


	static int a[] = new int[100000+5];
	static int MOD = 1000 * 1000 * 1000 + 7;
	public static void main(String[]args)throws Throwable
	{
		PriorityQueue<Long> powers = new PriorityQueue();
		int x;
		long n;
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		n = new Integer(st.nextToken());
		x = new Integer(st.nextToken());
		st = new StringTokenizer(br.readLine());

		long sum = 0;
		for(int i = 0 ; i < n ; ++i)
		{

			a[i] = new Integer(st.nextToken());
			sum += a[i];
		}
		for(int i = 0 ; i < n ; i++)
		{
			powers.add(sum - a[i]);
		}
		long sol = powers.poll();
		int cnt = 1;

		while(true)
		{

			while(!powers.isEmpty())
			{
				long out_p = powers.poll() - sol;
				if(out_p == 0) ++cnt;
				else 
				{
					powers.add(out_p + sol);
					break;
				}
			}
			//sol += (v-sol);
			
			//System.err.println(sol);
			if(cnt == 0 || cnt % x != 0) break;

++sol;
			cnt /= x;

		}
		sol = Math.min(sum,sol);
		BigInteger res = new BigInteger(x+"");
		res = res.modPow(new BigInteger(sol+""), new BigInteger(MOD+""));

		System.out.println(res);


	}
}
