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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.HashMap;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author NRK7
 */
public class Main {
	public static void main (String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader (inputStream);
		PrintWriter out = new PrintWriter (outputStream);
		TaskC solver = new TaskC ();
		solver.solve (1, in, out);
		out.close ();
	}
	
	static class TaskC {
		private static long modpowx (int x, long n) {
			long res = 1, os = x;
			while (n>0) {
				if ((n&1) != 0) res = (res*os)%InputReader.inf;
				os = (os*os)%InputReader.inf;
				n >>= 1;
			}
			return res;
		}
		
		public void solve (int testNumber, InputReader in, PrintWriter out) {
/* Hope it is in my range. 1: First of all find amax and then sum(x^(amax-a[i])), find gcd between...
 * multiply this with cs(a)/amax. Only challenge in this approach is finding sum(x^) is impossible as
 * (1e9^1e9). To find gcd directly, one way may be to take common the lowest power of amax in x^(amax-a)
 * But another thing to concern is 1+1+1+1+1+1+3+9+27 at numerator (x=3) ==> if cnt of any power of x is
 * divisor of x then it can be joined with cnt (here (6 ==> 3*2, 3*(2+1) ==> 9*1, 9*(1+1) 9 is os, as
 * 2 is not divisor of x, so it can't be joined with later ones and it is gcd of this part)). I have no
 * idea why this solution failed in 2 test cases, when the array is sorted no need to use map to cnt[]*/
			int n = in.nextInt (), x = in.nextInt ();
			int[] a = new int[n+5];
			long cs = 0;
			for (int i = 0; i<n; i++) {
				a[i] = in.nextInt ();
				cs += a[i];
			}
			HashMap<Long, Integer> cntpx = new HashMap<> ();
			for (int i = 0; i<n; i++) {
				cntpx.put (cs-a[i], cntpx.getOrDefault (cs-a[i], 0)+1);
			}
			long os = -1;
			int op = 0;
			for (long i = cs-a[n-1]; i<cs+2; i++) {
				op = cntpx.getOrDefault (i, 0);
				if (op%x == 0) {
					cntpx.put ((long) i+1, cntpx.getOrDefault (i+1, 0)+op/x);
				}
				else {
					os = i;
					break;//that's why it won't last for more than 1e5 times
				}
			}
			if (os == -1) {
				os = cs+1;
				while (op%x == 0) {
					op /= x;
					os++;
				}
			}
			os = Math.min (os, cs);
			os = modpowx (x, os);
			out.println (os);
		}
		
	}
	
	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;
		public static final int inf = 1000000007;
		
		public InputReader (InputStream stream) {
			reader = new BufferedReader (new InputStreamReader (stream));
			tokenizer = null;
		}
		
		public String next () {
			while (tokenizer == null || !tokenizer.hasMoreTokens ()) {
				try {
					String str;
					if ((str = reader.readLine ()) != null) tokenizer = new StringTokenizer (str);
					else return null;//to detect eof
				} catch (IOException e) {
					throw new RuntimeException (e);
				}
			}
			return tokenizer.nextToken ();
		}
		
		public int nextInt () {
			return Integer.parseInt (next ());
		}
		
	}
}

