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.io.*;

public class C {
	FastScanner in;
	PrintWriter out;

	ArrayList<Integer>[] g;
	long[][] cnt;
	int n;
	int k;
	long ans;
	
	void dfs1(int v, int p) {
		cnt[v][0] = 1;
		for (int u : g[v]) {
			if (u != p) {
				dfs1(u, v);
				for (int i = 0; i < k; i++) {
					cnt[v][(i + 1) % k] += cnt[u][i];
				}
			}
		}
	}
	
	void dfs(int v, int p, long[] cntUp) {
		long[] cntV = new long[k];
		for (int i = 0; i < k; i++) {
			cntV[i] += cntUp[i];
			for (int u : g[v]) {
				if (u != p) {
					cntV[(i + 1) % k] += cnt[u][i];
				}
			}
		}
		long[] newCntUp = new long[k];
		for (int u : g[v]) {
			if (u != p) {
				long sumSub = 0;
				for (int i = 0; i < k; i++) {
					sumSub += cnt[u][i];
				}
				ans += cnt[u][0] * (n - sumSub) + (cntV[0] - cnt[u][k - 1]) * sumSub;
				for (int i = 0; i < k; i++) {
					newCntUp[i] = cntV[(i + k - 1) % k] - cnt[u][(i + k - 2) % k];
				}
				newCntUp[0] += 1;
				dfs(u, v, newCntUp);
			}
		}
	}
	
	public void solve() throws IOException {
		n = in.nextInt();
		k = in.nextInt();
		cnt = new long[n][k];
		g = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
		}
		for (int i = 0; i < n - 1; i++) {
			int a = in.nextInt() - 1, b = in.nextInt() - 1;
			g[a].add(b);
			g[b].add(a);
		}
		long[] cntUp = new long[k];
		dfs1(0, -1);
		cntUp[0] = 1;
		dfs(0, -1, cntUp);
		out.println(ans / 2);
	}

	public void run() {
		try {
			in = new FastScanner();
			out = new PrintWriter(System.out);

			solve();

			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

	public static void main(String[] arg) {
		new C().run();
	}
}