CF771C：Bear and Tree Jumps（树形dp & 树上距离和）

C. Bear and Tree Jumps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 21 21 32 42 54 6

output

20

input

13 31 23 24 25 23 610 66 76 135 85 99 1111 12

output

114

input

3 52 13 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.

题意：一棵树，人每次最多跳K距离，即从i跳到j需要(dis(i,j)-1)/K+1次，问跳完所有点对的次数之和。

思路：树形dp，dp[i][j]表示i点到根节点的距离%k=j的节点数，这里的i是由底往上递归，所有节点的距离之和+所有节点差多少能被k整除的和，再除以k就是答案。

# include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 2e5+30;vector<int>v[maxn];int n, k;LL ans = 0, dp[maxn][5],h[maxn];void dfs(int cur, int pre, int d){    h[cur] = 1;    dp[cur][d%k] = 1;    for(auto to :v[cur])    {        if(to == pre) continue;        dfs(to, cur, d+1);        for(int i=0; i<k; ++i)        {            for(int j=0; j<k; ++j)            {                int x = (i+j-d*2)%k;//经过以cur为根节点的两点LCA。                int y = (k-x)%k;//差多少才能被k整除。                ans += y*dp[cur][i]*dp[to][j];            }        }        for(int i=0; i<k; ++i)            dp[cur][i] += dp[to][i];        ans += (n-h[to])*h[to];//计算树的两两距离（相当于累加这些边的贡献次数）。        h[cur] += h[to];    }}int main(){    int a, b;    scanf("%d%d",&n,&k);    for(int i=1; i<n; ++i)    {        scanf("%d%d",&a,&b);        v[a].push_back(b);        v[b].push_back(a);    }    dfs(1,0,0);    printf("%lld\n",ans/k);    return 0;}