Codeforce 862B Mahmoud and Ehab and the bipartiteness （二分图染色 + 链式前向星）

Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.

A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (u, v) that belongs to the graph, u and v belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.

Dr. Evil gave Mahmoud and Ehab a tree consisting of n nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?

A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .

Input

The first line of input contains an integer n — the number of nodes in the tree (1 ≤ n ≤ 10⁵).

The next n - 1 lines contain integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the description of the edges of the tree.

It's guaranteed that the given graph is a tree.

Output

Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.

Example

Input

31 21 3

Output

0

Input

51 22 33 44 5

Output

2

e题意：

给你 n 代表n个节点。

给你n-1条边

问你再最多再添加多少边使得二分图的性质成立；

二分图简介

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 200005struct data{    int to,next;} Line[N];int cnt = 0;int Color[N];int head[N];void add(int a,int b){    Line[cnt].to=b;    Line[cnt].next=head[a];    head[a]=cnt++;}void Add_Line(int a,int b){    add(a,b);    add(b,a);}int dfs(int x,int color){    Color[x]=color;    for(int i=head[x]; i!=-1; i=Line[i].next)    {        if(Color[Line[i].to]==-1)            dfs(Line[i].to,!color);    }}int main(){    int n;    while(~scanf("%d",&n))    {        memset(head,-1,sizeof(head));        memset(Color,-1,sizeof(Color));        for(int i=1; i<n; i++)        {            int x,y;            scanf("%d%d",&x,&y);            Add_Line(x,y);        }        dfs(1,0);        long long res=0;        for(int i=1; i<=n; i++)        {            res+=(Color[i]);        }        res = res*(n-res);        printf("%lld\n",res-(n-1));    }}