Description

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 off(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 most2. 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.

题意:给出一棵树,和最多跳K个数字,f(s,t)表示从s到t需要跳的最少次数,问那么一棵树每两个点跳的次数之和是多少?

解法:如果只跳一次,那每个点经历的次数为这个点的子树节点个数*(n-这个点的子树节点个数),那么K>=2的情况,对于任意两个点的x->y 距离为 深度[x]+深度[y]-2*最近公共祖先深度[z]

dp[v][d%k]表示v为节点开始,深度为d%k的个数,比如2为节点开始,深度为1的有4和5

4在2的子树内,4开始深度为1点为6,我们更新到dp[2][1]内,就是dp[2][1]+=dp[4][1]

sum[v]表示v的子树节点个数

最后答案为ans/n,比如距离为5,最多跳3步,其实是跳(5+1)/3=2次就好了,注释也有解释

(题解好难懂啊,我也不知道说清楚了没)

 #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=;
ll dp[maxn][],sum[maxn];
vector<int>q[maxn];
ll vis[maxn];
ll cnt;
ll n,k;
void dfs(int v,int d,int fa)
{
dp[v][d%k]=sum[v]=;
//当前子节点就v一个
for(int i=;i<q[v].size();i++)
{
// cout<<v<<endl;
int pos=q[v][i];
if(pos==fa) continue;
dfs(pos,d+,v);
for(int x=;x<k;x++)
{
for(int y=;y<k;y++)
{
int ans=((x+y)%k-(d*)%k+k)%k;
//ans为缺少部分,比如5跳3,少了两步
cnt+=((k-ans)%k)*dp[v][x]*dp[pos][y];
//少了两步,为了达成三步,必须多走一步,所以为k-ans,每个点多走k-ans步,相乘
}
}
for(int x=;x<k;x++)
{
dp[v][x]+=dp[pos][x];
//子节点记录部分更新到父结点
}
cnt+=sum[pos]*(n-sum[pos]);
//讨论k=1的情况,种数==为pos节点包含子节点*以外的节点
sum[v]+=sum[pos];
//将pos包含节点个数更新到父结点 }
}
int main()
{
cin>>n>>k;
for(int i=;i<n-;i++)
{
int u,v;
cin>>u>>v;
q[u].push_back(v);
q[v].push_back(u);
}
dfs(,,);
cout<<cnt/k<<endl;
return ;
}

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