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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF359C extends PrintWriter {
	CF359C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF359C o = new CF359C(); o.main(); o.flush();
	}

	static final int MD = 1000000007;
	long power(long a, int k) {
		long p = 1;
		while (k > 0) {
			if ((k & 1) == 1)
				p = p * a % MD;
			a = a * a % MD;
			k >>= 1;
		}
		return p;
	}
	void main() {
		int n = sc.nextInt();
		int x = sc.nextInt();
		long[] aa = new long[n + 1]; long s = 0;
		for (int i = 0; i < n; i++)
			s += aa[i] = sc.nextInt();
		for (int i = 0; i < n; i++)
			aa[i] = s - aa[i];
		aa[n] = s + 20;
		Arrays.sort(aa);
		long a_ = -1;
out:
		for (int i = 0, j, k = 0; i <= n; i = j) {
			long a = aa[i];
			if (k > 0)
				while (a_ < a) {
					if (k % x > 0)
						break out;
					k /= x;
					a_++;
				}
			j = i + 1;
			while (j < n && aa[j] == a)
				j++;
			k += j - i;
			a_ = a;
		}
		println(power(x, (int) (Math.min(a_, s) % (MD - 1))));
	}
}
