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.
// package CF;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;


public class A {
	static int n, k, sub [], cnt[][];
	static long ans;
	static ArrayList<Integer> [] adj;
	static void dfs(int u, int p, int d){
		sub[u] = 1;
		cnt[u][d%k] = 1;
		for(int v:adj[u])
			if(v != p){
				dfs(v, u, d+1);
				for (int i = 0; i < k; i++) {
					for (int j = 0; j < k; j++) {
						ans += sub(k, sub(i+j, 2*d))*cnt[u][i]*cnt[v][j];
					}
				}
				sub[u] += sub[v];
				for (int i = 0; i < k; i++) {
					cnt[u][i] += cnt[v][i];
				}
			}
		ans += 1l*sub[u]*(n-sub[u]);
	}
	static long sub(long a, long b){
		return ((a - b) % k + k) % k;
	}
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		n = sc.nextInt(); k = sc.nextInt();
		ans = 0;
		adj = new ArrayList[n];
		for (int i = 0; i < adj.length; i++) {
			adj[i] = new ArrayList<>();
		}
		for (int i = 0; i < n-1; i++) {
			int u = sc.nextInt(), v = sc.nextInt();
			u--; v--;
			adj[u].add(v);
			adj[v].add(u);
		}
		sub = new int[n];
		cnt = new int[n][k];
		dfs(0, -1, 0);
		out.println(ans/k);
		out.flush();
		out.close();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader fileReader) {
			br = new BufferedReader(fileReader);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}