A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them.

Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.

Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.

For a pair of vertices (s, t) we define f(s, t) as the minimum number of jumps Limak needs to get from s to t. Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Input

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) — the number of vertices in the tree and the maximum allowed jump distance respectively.

The next n - 1 lines describe edges in the tree. The i-th of those lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) — the indices on vertices connected with i-th edge.

It's guaranteed that the given edges form a tree.

Output

Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Examples

Input

6 2
1 2
1 3
2 4
2 5
4 6


Output

20


Input

13 3
1 2
3 2
4 2
5 2
3 6
10 6
6 7
6 13
5 8
5 9
9 11
11 12


Output

114


Input

3 5
2 1
3 1


Output

3

Note

In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at most 2. For example, from the vertex 5 he can jump to any of vertices: 1, 2 and 4 (well, he can also jump to the vertex 5 itself).

<image>

There are <image> pairs of vertices (s, t) such that s < t. For 5 of those pairs Limak would need two jumps: (1, 6), (3, 4), (3, 5), (3, 6), (5, 6). For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.

In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t), so the answer is 3·1 = 3.


import java.util.*;
import java.lang.*;
import java.io.*;
 
 

public class D
{
    static long answer = 0;
    static int n = 0;
    static int k = 0;
    static ArrayList<Integer>[] adjacencyList = new ArrayList[200000];
    static int[] childs = new int[200000];
    static int[][] f = new int[200000][5];

    public static void rec(int node, int parent)
    {
        f[node][0] = 1;

        int c = 1;
        for (int ch=0; ch<adjacencyList[node].size(); ch++)
        {
            int child = adjacencyList[node].get(ch);
            if (child == parent)
                continue;

            rec(child, node);
            answer += ((long)childs[child] * (n - childs[child]));
            c += childs[child];

            for (int i=0; i<k; i++)
                for (int j=0; j<k; j++)
                {
                    int needs = (k - ((i + 1 + j) % k)) % k;
                    answer += ((long)needs * f[node][j] * f[child][i]);
                }

            for (int i=0; i<k; i++)
                f[node][(i + 1) % k] += f[child][i];
        }

        childs[node] = c;
    }

    public static void main(String[] args)
    {
            Scanner br = new Scanner(System.in);
            //Scanner br = new Scanner();

            n = br.nextInt();
            k = br.nextInt();

            for (int i=0; i<n; i++)
                adjacencyList[i] = new ArrayList<Integer>();

            for (int i=0; i<n - 1; i++)
            {
                int a = br.nextInt();
                int b = br.nextInt();
                adjacencyList[a - 1].add(b - 1);
                adjacencyList[b - 1].add(a - 1);
            }

            rec(0, -1);
            System.out.println(answer/k);
    }
}

