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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;

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

@SuppressWarnings("unused")
public class Main {
	public static void main(String[] args) throws IOException {
		new Main().run();		
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");

	private void run() throws IOException {
		if (new File("input.txt").exists())
			in = new BufferedReader(new FileReader("input.txt"));
		else
			in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		in.close();
		out.close();
	}
	
	int N = 1000 * 1000;
	int[] head = new int[N];
	int[] next = new int[N];
	int[] to = new int[N];
	int cnt = 0;
	int n = 0;
	int k = 0;
	boolean[] used = new boolean[N];
	long ans = 0;
	long[][] a;
	long[][] b;
	long[] ca;
	long[] cb;
	
	private void solve() throws IOException {
		
		n = nextInt();
		k = nextInt();
		fill(head, -1);
		for (int i = 0; i < n - 1; i++) {
			add(nextInt() - 1, nextInt() - 1);
		}
		a = new long[k][n];
		b = new long[k][n];
		ca = new long[k];
		cb = new long[k];
		rec(0);
		out.println(ans);
//		for (int i = 0; i < n; i++) {
//			for (int j = 0; j < k; j++)
//				System.err.print(a[j][i] + "," + b[j][i] + " ");
//			System.err.println();
//		}
	}
	
	void rec(int u) {
		used[u] = true;
//		System.err.println("+" + u);
		for (int i = head[u]; i != -1; i = next[i]) {
			int v = to[i];
			if (used[v])
				continue;
			rec(v);
			for (int l = 0; l < k; l++) {
				for (int r = 0; r < k; r++) {
					long x = a[l][u] * b[r][v];
					x += b[l][u] * a[r][v];
					x += ((l + r + 2L) / k + (((l + r + 2) % k == 0) ? 0 : 1)) * b[l][u] * b[r][v];
					ans += x;
				}
			}
			for (int l = 0; l < k; l++) {
				a[l][u] += a[l][v];
				b[l][u] += b[l][v];
			}
		}
		for (int i = 0; i < k; i++) {
			ca[i] = a[i][u];
			cb[i] = b[i][u];
		}
//		for (int l = 0; l < k; l++) {
//			System.err.println(ca[l] + " " + cb[l]);
//		}
		for (int l = 0; l < k - 1; l++) {
			ans += ca[l] + cb[l];
			a[l + 1][u] = ca[l];
			b[l + 1][u] = cb[l];
		}
		ans += ca[k - 1] + cb[k - 1];
		a[0][u] = ca[k - 1] + cb[k - 1];
		b[0][u] = cb[k - 1] + 1;
//		System.err.println("-" + u);
	}
	
	void add(int u, int v) {
		to[cnt] = v;
		next[cnt] = head[u];
		head[u] = cnt++;
		to[cnt] = u;
		next[cnt] = head[v];
		head[v] = cnt++;
	}
	
	void chk(boolean b) {
		if (b)
			return;
		System.out.println(new Error().getStackTrace()[1]);
		exit(999);
	}
	void deb(String fmt, Object... args) {
		System.out.printf(Locale.US, fmt + "%n", args);
	}
	String nextToken() throws IOException {
		while (!st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}
	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null)
				return true;
			st = new StringTokenizer(s);
		}
		return false;
	}
}
