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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class C {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        long x = Long.parseLong(st.nextToken());

        st = new StringTokenizer(br.readLine());
        long[] nums = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            nums[i] = Long.parseLong(st.nextToken());
            sum += nums[i];
        }

        PriorityQueue<Long> pow = new PriorityQueue<Long>();
        for (int i = 0; i < n; i++)
            pow.add(sum - nums[i]);

        while (true) {
            long b = pow.peek();
            int cnt = 0;
            for (int i = 0; i < x && !pow.isEmpty(); i++) {
                if (pow.poll() == b)
                    cnt++;
                else
                    break;
            }

            if (cnt == x)
                pow.add(b + 1);
            else {
                pow.add(b);
                break;
            }
        }

        long ans = Math.min(pow.peek(), sum);
        BigInteger big = BigInteger.valueOf(x).modPow(BigInteger.valueOf(ans), BigInteger.valueOf(1000000007));
        System.out.println(big.toString());
    }
}