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 c {
public static void main(String[] args)
{
    Scanner input = new Scanner(System.in);
    int n = input.nextInt(), x = input.nextInt();
    long mod = 1000000007;
    int[] data = new int[n];
    for(int i = 0; i<n; i++) data[i] = input.nextInt();
    long sum = 0;
    for(int i =0; i<n; i++) sum += data[i];
    int count = 1;
    long res = 2000000000;
    for(int i =n-2; i>=0; i--)
    {
        if(data[i] == data[i+1]) count++;
        else
        {
            int q = count / x;
            int r = count % x;
            if(r > 0)
            {
                res = data[i+1];
                break;
            }
            boolean found = false;
            int qq = q;
            while(qq > 0)
            {
                if(data[i] == -1 + data[i+1])
                {
                    found = true;
                    count = 1+qq;
                    break;
                }
                if(qq%x != 0)
                {
                    res = data[i+1]-1;
                    break;
                }
                qq /= x;
                data[i+1]--;
            }
            if(res != 2000000000) break;
            if(!found)
            {
                count = 1;
                if(q%x != 0)
                {
                    res = -1+data[i+1];
                    break;
                }
            }
        }
    }
    if(res == 2000000000)
    {
        res = data[0];
        while(count%x == 0)
        {
            res--;
            count = count/x;
        }
    }
    System.out.println(pow(x, Math.min(sum,sum-res)));
}
static long mod = 1000000007;
static long pow(long x, long p)
{
    if(p==0) return 1;
    if((p&1) == 0)
    {
        long sqrt = pow(x, p/2);
        return (sqrt*sqrt)%mod;
    }
    return (x * pow(x, p-1))%mod;
}
}
