Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies.

On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection ofn points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by apath (a sequence of one or more edges).

To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!

The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.

Print one integer – the number of lifelines in the tree.

In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.

分析：长度为2的路的个数；

代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#include <ext/rope>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define vi vector<int>

#define pii pair<int,int>

#define mod 1000000007

#define inf 0x3f3f3f3f

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

const int maxn=1e4+;

const int dis[][]={{,},{-,},{,-},{,}};

using namespace std;

using namespace __gnu_cxx;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,vis[maxn],cnt;

vi a[maxn];

void dfs(int p)

{

    for(int x:a[p])for(int y:a[x])if(!vis[y])cnt++;

}

int main()

{

    int i,j,k,t;

    scanf("%d",&n);

    rep(i,,n-)scanf("%d%d",&j,&k),a[j].pb(k),a[k].pb(j);

    rep(i,,n)vis[i]=,dfs(i);

    printf("%d\n",cnt);

    //system ("pause");

    return ;

}