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.util.*;
import java.math.*;

public class Cf359c {
    static long M = (long)1e9 + 7;
    static int n;
    static long x;
    static long[] a = new long[101000];
    static long sm = 0;

    static long qp(long n, long m) {
        long b = 1;
        while (m != 0) {
            if (m % 2 == 1) b = b*n%M;
            m >>= 1;
            n = n*n%M;
        }
        return b;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt(); x = sc.nextLong();
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
            sm += a[i];
        }
        for (int i = 0; i < n; i++) a[i] = sm - a[i];
        a[n] = sm; n++;
        Arrays.sort(a, 0, n);
        long pre = a[0], ca = 0;
        long ans = sm;
        boolean fl = false;
        for (int i = 0; i < n; i++, ca++) {
            if (a[i] == pre) continue;
            while (pre != a[i]) {
                if (ca % x == 0) {
                    ca /= x;
                    pre++;
                } else {
                    fl = true;
                    ans = pre;
                    break;
                }
            }
            if (fl) break;
        }
        System.out.println(qp(x, ans));
    }
}
