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 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.

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 200005using 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;}