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.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;


// author: KAN
// editorial

public class Flab {

	static List<List<Integer>> adj;
	static int v;
	
	// coutn at cur vert that mods k
	static long dp[][] = new long[200000][6];
	static long tot[] = new long[200000];
	
	static int k;
	
	static long ans = 0;

	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(System.in);
		
		v = in.nextInt();
		k = in.nextInt();
		
		adj = new ArrayList<>();
		
		for(int i=0; i<v; i++) {
			adj.add(new ArrayList<>());
		}
		
		for(int i=0; i<v-1; i++) {
			int x = in.nextInt()-1;
			int y = in.nextInt()-1;
			
			adj.get(x).add(y);
			adj.get(y).add(x);
		}
		
		dfs(0, -1, 0);
		
		System.out.println(ans/k);

	}

	private static void dfs(int i, int par, int dep) {
		tot[i] = 1;
		dp[i][dep%k] = 1;
		
		for(int n : adj.get(i)) {
			if(n == par) {
				continue;
			}
			
			dfs(n, i, dep+1);
			tot[i] += tot[n];
			
			for(int u = 0; u<k; u++) {
				for(int v=0; v<k; v++) {
					
					//finding lengths distance between u and v vertices mod k
					long dist = (((u+v - (2*dep))%k)+k)%k;

					
					long add =  dist == 0 ? 0 : k - dist;
					

					ans += (add*dp[n][u]*dp[i][v]);
					
				}
			}
			
			
			for(int g=0; g<k; g++) {
				dp[i][g] += dp[n][g];
			}
			
			ans += (tot[n] * (v - tot[n]));
			
		}
		
	}


}
