#include <bits/stdc++.h>
using namespace std;

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

typedef long long LL;

int uf[1000];

int find(int a){
  if(uf[a]==a) return a;
  else return uf[a]=find(uf[a]);
}

void uni(int a,int b){
  find(a);
  find(b);
  uf[uf[a]]=uf[b];
  return;
}

struct node{
  int s;
  int t;
  int c;
  bool operator<(const struct node& another) const{
    return c<another.c;
  }
};

int main(){
  int n,m;
  while(1){
    cin >> n >> m;
    if(n==0) return 0;
    vector<struct node> e(m);
    int s,t,c;
    for(int i=0;i<n;i++){
      uf[i]=i;
    }
    for(int i=0;i<m;i++){
      cin >> e[i].s >> e[i].t >> e[i].c;
      e[i].s--;
      e[i].t--;
    }
    sort(e.begin(),e.end());
    int count=0;
    vector<int> cost(n-1);
    for(int i=0;i<m;i++){
      if(find(e[i].s)!=find(e[i].t)){
        uni(e[i].s,e[i].t);
        cost[count]=e[i].c;
        count++;
      }
      if(count==n-1){
        cout << cost[n/2-1] << endl;
        break;
      }
    }
  }
  return 0;
}


### Prompt

Please create a solution in cpp 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 <bits/stdc++.h>
using namespace std;

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

typedef long long LL;

int uf[1000];

int find(int a){
  if(uf[a]==a) return a;
  else return uf[a]=find(uf[a]);
}

void uni(int a,int b){
  find(a);
  find(b);
  uf[uf[a]]=uf[b];
  return;
}

struct node{
  int s;
  int t;
  int c;
  bool operator<(const struct node& another) const{
    return c<another.c;
  }
};

int main(){
  int n,m;
  while(1){
    cin >> n >> m;
    if(n==0) return 0;
    vector<struct node> e(m);
    int s,t,c;
    for(int i=0;i<n;i++){
      uf[i]=i;
    }
    for(int i=0;i<m;i++){
      cin >> e[i].s >> e[i].t >> e[i].c;
      e[i].s--;
      e[i].t--;
    }
    sort(e.begin(),e.end());
    int count=0;
    vector<int> cost(n-1);
    for(int i=0;i<m;i++){
      if(find(e[i].s)!=find(e[i].t)){
        uni(e[i].s,e[i].t);
        cost[count]=e[i].c;
        count++;
      }
      if(count==n-1){
        cout << cost[n/2-1] << endl;
        break;
      }
    }
  }
  return 0;
}


```