Codeforces Round #405 Div. 1 B. Bear and Tree Jumps

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).

There are  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.

分析:一棵子树的所有路径可以分为过根和不过根节点,这题特殊在一步可以走k(k <= 5)条边,所以我们把所有路径按照模k划分为k类然后每次dfs时讨论一下就可以了,复杂度O(n*k^2).

#include <bits/stdc++.h>
#define MOD 10000007
#define N 200005
using namespace std;
int n,k,u,v;
long long ans,tf[5],tg[5],f[N][5],g[N][5];
vector<int> G[N];
void dfs(int u,int fa)
{
	g[u][0]++;
	for(int i = 0;i < G[u].size();i++)
	{
		int v = G[u][i];
		if(v != fa) 
		{
			dfs(v,u); 
			for(int j = 0;j < k;j++)
			{
				tf[(j+1) % k] = f[v][j] + ((j+1)/k)*g[v][j];
				tg[(j+1) % k] = g[v][j]; 	
			}	
			for(int j = 0;j < k;j++)
			 for(int j2 = 0;j2 < k;j2++)
			  ans += tf[j]*g[u][j2]+tg[j]*f[u][j2]+((bool)((j+j2)%k) + (j+j2)/k)*g[u][j2]*tg[j]; 
			for(int j = 0;j < k;j++) 
			{
				f[u][j] += tf[j];
				g[u][j] += tg[j];
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i = 1;i < n;i++)
	{
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1,1);
	cout<<ans<<endl;
}