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.util.*;

public class P359C {
    // a ^ b mod n
    private static int modExp(int a, int b, int n) {
        long c = 0;
        long d = 1;
        ArrayList<Integer> bRep = new ArrayList<Integer>();
        int bp = b;
        while (bp > 0) {
            bRep.add(bp & 1);
            bp = bp >> 1;
        }
        for (int i = bRep.size() - 1; i >= 0; i--) {
            c *= 2;
            d = (d * d) % n;
            if (bRep.get(i) == 1) {
                c += 1;
                d = (d * a) % n;
            }
        }
        
        return (int)d;
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int x = sc.nextInt();
        long[] a  = new long[N];
        long[] ap = new long[N];
        long aSum = 0;
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
            aSum += a[i];
        }
        for (int i = 0; i < N; i++) {
            ap[i] = aSum - a[i];
        }
        Arrays.sort(ap);
        TreeMap<Long, Integer> digits = new TreeMap<Long, Integer>();
        for (long p: ap) {
            if (!digits.containsKey(p)) {
                digits.put(p, 0);
            }
            int d = digits.get(p);
            assert d < x;
            if (d == x - 1) {
                long q = p;
                while (digits.containsKey(q) && digits.get(q) == x - 1) {
                    digits.remove(q++);
                }
                if (!digits.containsKey(q)) {
                    digits.put(q, 0);
                }
                digits.put(q, digits.get(q) + 1);
            } else {
                digits.put(p, d + 1);
            }
        }
        
        long y = digits.firstKey();
        
        final int MVAL = 1000000007;
        
        int result = 0;
        
        if (x % MVAL == 0) {
            result = (Math.min(y, aSum) == 0 ? 1 : 0);
        } else {
            int xp = x % MVAL;
            int yp;
            if (y >= aSum) {
                yp = (int)(aSum % (MVAL - 1));
            } else {
                yp = (int)(y % (MVAL - 1));
            }
            result = modExp(xp, yp, MVAL);
        }
        
        System.out.println(result);
    }
}
