#include<utility>    //pair
#include<vector>
#include<stdio.h>
#include<string>
#include<iostream>
#include<algorithm> //find的&#22836;文件  string用find的&#26102;候会返回字符串的首地址
#include<iterator>
#include<string.h>
#include<deque>   //双向添加数字的数&#32452;  可以&#21024;除或者添加第一个
#include<queue>
#include<set>      //&#24635;&#35273;得没有什&#20040;用&#22788;  无重&#22797;的
#include<list>    //不支持随机&#35775;&#38382;  即缺少下&#26631;
#include<map>       //数据&#31867;型很方便 但是会自己排好序  而且不会有重&#22797;的元素
using namespace std;
int cost[1005][1005];
int mincost[1005];
bool used[1005];
const int INF=10000000;
int d[1005],flag;
int main()
{
    int V,m,a,b,c,t;
    while(cin>>V>>m)
    {
        for(int i=0;i<1005;i++)
            for(int j=0;j<1005;j++)
                cost[i][j]=INF;

        if(V==m&&m==0)
            break;
        flag=0;
        for(int i=0;i<m;i++)
        {

            cin>>a>>b>>c;
            cost[a][b]=c;
            cost[b][a]=c;

        }
        for(int i=1;i<=V;i++)
        {
            mincost[i]=INF;
            used[i]=false;
        }
        mincost[1]=0;
        while(true)
        {
            int v=-1;
            for(int u=1;u<=V;u++)
            {
                if(!used[u]&&(v==-1||mincost[u]<mincost[v]))
                    v=u;
            }
            if(v==-1)
                break;
            used[v]=true;
            d[flag]=mincost[v];flag++;
            for(int u=1;u<=V;u++)
            {
                mincost[u]=min(mincost[u],cost[v][u]);
            }

        }
        sort(d,d+flag);
        cout<<d[V/2]<<endl;

    }

    return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

Problem Statement

You are given a connected undirected graph which has even numbers of nodes. A connected graph is a graph in which all nodes are connected directly or indirectly by edges.

Your task is to find a spanning tree whose median value of edges' costs is minimum. A spanning tree of a graph means that a tree which contains all nodes of the graph.



Input

The input consists of multiple datasets.

The format of each dataset is as follows.


n m
s_1 t_1 c_1
...
s_m t_m c_m


The first line contains an even number n (2 \leq n \leq 1,000) and an integer m (n-1 \leq m \leq 10,000). n is the nubmer of nodes and m is the number of edges in the graph.

Then m lines follow, each of which contains s_i (1 \leq s_i \leq n), t_i (1 \leq s_i \leq n, t_i \neq s_i) and c_i (1 \leq c_i \leq 1,000). This means there is an edge between the nodes s_i and t_i and its cost is c_i. There is no more than one edge which connects s_i and t_i.

The input terminates when n=0 and m=0. Your program must not output anything for this case.

Output

Print the median value in a line for each dataset.

Example

Input

2 1
1 2 5
4 6
1 2 1
1 3 2
1 4 3
2 3 4
2 4 5
3 4 6
8 17
1 4 767
3 1 609
8 3 426
6 5 972
8 1 607
6 4 51
5 1 683
3 6 451
3 4 630
8 7 912
3 7 43
4 7 421
3 5 582
8 4 538
5 7 832
1 6 345
8 2 608
0 0


Output

5
2
421

### Response

```cpp
#include<utility>    //pair
#include<vector>
#include<stdio.h>
#include<string>
#include<iostream>
#include<algorithm> //find的&#22836;文件  string用find的&#26102;候会返回字符串的首地址
#include<iterator>
#include<string.h>
#include<deque>   //双向添加数字的数&#32452;  可以&#21024;除或者添加第一个
#include<queue>
#include<set>      //&#24635;&#35273;得没有什&#20040;用&#22788;  无重&#22797;的
#include<list>    //不支持随机&#35775;&#38382;  即缺少下&#26631;
#include<map>       //数据&#31867;型很方便 但是会自己排好序  而且不会有重&#22797;的元素
using namespace std;
int cost[1005][1005];
int mincost[1005];
bool used[1005];
const int INF=10000000;
int d[1005],flag;
int main()
{
    int V,m,a,b,c,t;
    while(cin>>V>>m)
    {
        for(int i=0;i<1005;i++)
            for(int j=0;j<1005;j++)
                cost[i][j]=INF;

        if(V==m&&m==0)
            break;
        flag=0;
        for(int i=0;i<m;i++)
        {

            cin>>a>>b>>c;
            cost[a][b]=c;
            cost[b][a]=c;

        }
        for(int i=1;i<=V;i++)
        {
            mincost[i]=INF;
            used[i]=false;
        }
        mincost[1]=0;
        while(true)
        {
            int v=-1;
            for(int u=1;u<=V;u++)
            {
                if(!used[u]&&(v==-1||mincost[u]<mincost[v]))
                    v=u;
            }
            if(v==-1)
                break;
            used[v]=true;
            d[flag]=mincost[v];flag++;
            for(int u=1;u<=V;u++)
            {
                mincost[u]=min(mincost[u],cost[v][u]);
            }

        }
        sort(d,d+flag);
        cout<<d[V/2]<<endl;

    }

    return 0;
}
```