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.Scanner;
import java.util.Arrays;
import java.util.Vector;

public class B {
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        long sum = 0;
        
        long n = in.nextInt();
        long x = in.nextInt();
        long[] a = new long [(int)n];
        long[] s = new long [(int)n];
        for(int i = 0; i < n; i++) { a[i] = in.nextInt(); sum += a[i]; }
        for(int i = 0; i < n; i++) { s[i] = sum - a[i]; }
        Arrays.sort(s);
        
        //System.out.println("here");
        
        long l = 0, r = n-1;
        long ans = s[0];
        while(true) {
            //System.out.println("here");
            long mark = s[(int)l];
            long cnt = 0;
            while(cnt+l <= r && s[(int)(cnt+l)] == s[(int)l]) cnt++;
            l += cnt;
            //System.out.println("here2");
            if(cnt % x == 0) {
                for(long i = 0; i < cnt/x; i++) s[(int)--l] = mark+1;
                ans++;
            }
            else break;
        }
        
        ans = Math.min(ans, sum);
        
        System.out.println(mPow(x, ans));
        
        in.close();
    }
    
    public static long mPow(long b, long p) {
        long ret = 1;
        long mod = 1000000007;
        while(p != 0) {
            if((p&1) == 1) { ret *= b; ret %= mod; }
            b = b*b;
            b %= mod;
            p >>= 1;
        }
        return ret;
    }
    
    public static long gcd(long a, long b) {
        if(a<0) return gcd(-a, b);
        if(b<0) return gcd(a, -b);
        return (b==0)?a:gcd(b,a%b);
    }
    public static long lcm(long a, long b) {
        if(a<0) return lcm(-a, b);
        if(b<0) return lcm(a, -b);
        return a*(b/gcd(a,b));
    }
}
