Helvetic Coding Contest 2016 online mirror F1

Description

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 of npoints (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (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!

Input

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.

Output

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

Examples

input

4
1 2
1 3
1 4

output

3

input

5
1 2
2 3
3 4
3 5

output

4

Note

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

我们可以dfs遍历，但只遍历两个点就return，每个点遍历一次，就是/2就好啦

#include<bits/stdc++.h>
using namespace std;
vector<int>q[100000];
int flag[100000];
int ans;
void dfs(int v,int sum)
{
    if(sum==2)
    {
        ans++;
        return;
    }
    if(flag[v])
    {
        return;
    }
    flag[v]=1;
    for(int i=0;i<q[v].size();i++)
    {
        if(!flag[q[v][i]])
        {
            dfs(q[v][i],sum+1);
        }
    }
}
int main()
{
    int n;
    int v,u;
    ans=0;
    cin>>n;
    for(int i=1;i<=n-1;i++)
    {
        cin>>v>>u;
        q[v].push_back(u);
        q[u].push_back(v);
    }
    for(int i=1;i<=n;i++)
    {
        memset(flag,0,sizeof(flag));
        dfs(i,0);
    }
    cout<<ans/2<<endl;
    return 0;
}