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

class uf{
public:
  uf();
  uf(int _n){
    init(_n);
  }
  void init(int _n){
    n = _n;
    rank.resize(n);
    parent.resize(n);
    for (int i = 0; i < n; i++) {
      rank[i] = 0;
      parent[i] = i;
    }
  }
  int find(int x){
    if(x == parent[x])return x;
    else return parent[x] = find(parent[x]);
  }
  
  void unite(int x, int y){
    x = find(x);
    y = find(y);
    if(x == y)return;
    if(rank[x] < rank[y]){
      parent[x] = y;
    }else{
      parent[y] = x;
      if(rank[x] == rank[y])rank[x]++;
    }
  }
  bool same(int x, int y){
    return find(x) == find(y);
  }
private:
  int n;
  vector<int> rank, parent;
};


struct edge{
  int u, v, cost;
  bool operator<(const edge &e)const{
    return cost < e.cost;
  }
};

vector<edge> kruskal(vector<edge> &es, int v){
  sort(es.begin(), es.end());
  uf u(v);
  vector<edge> res;
  for (int i = 0; i < es.size(); i++) {
    edge &e = es[i];
    if(!u.same(e.u, e.v)){
      u.unite(e.u, e.v);
      res.push_back(es[i]);
    }
  }
  return res;
}

int main(){
  int n, m;
  while(std::cin >> n >> m, n){
    vector<edge> ve(m);
    for (int i = 0; i < m; i++) {
      std::cin >> ve[i].u >> ve[i].v >> ve[i].cost;
      ve[i].u--;ve[i].v--;
    }
    vector<edge> ans = kruskal(ve, n);
    std::cout << ans[(n - 1)/2].cost << std::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<bits/stdc++.h>
using namespace std;

class uf{
public:
  uf();
  uf(int _n){
    init(_n);
  }
  void init(int _n){
    n = _n;
    rank.resize(n);
    parent.resize(n);
    for (int i = 0; i < n; i++) {
      rank[i] = 0;
      parent[i] = i;
    }
  }
  int find(int x){
    if(x == parent[x])return x;
    else return parent[x] = find(parent[x]);
  }
  
  void unite(int x, int y){
    x = find(x);
    y = find(y);
    if(x == y)return;
    if(rank[x] < rank[y]){
      parent[x] = y;
    }else{
      parent[y] = x;
      if(rank[x] == rank[y])rank[x]++;
    }
  }
  bool same(int x, int y){
    return find(x) == find(y);
  }
private:
  int n;
  vector<int> rank, parent;
};


struct edge{
  int u, v, cost;
  bool operator<(const edge &e)const{
    return cost < e.cost;
  }
};

vector<edge> kruskal(vector<edge> &es, int v){
  sort(es.begin(), es.end());
  uf u(v);
  vector<edge> res;
  for (int i = 0; i < es.size(); i++) {
    edge &e = es[i];
    if(!u.same(e.u, e.v)){
      u.unite(e.u, e.v);
      res.push_back(es[i]);
    }
  }
  return res;
}

int main(){
  int n, m;
  while(std::cin >> n >> m, n){
    vector<edge> ve(m);
    for (int i = 0; i < m; i++) {
      std::cin >> ve[i].u >> ve[i].v >> ve[i].cost;
      ve[i].u--;ve[i].v--;
    }
    vector<edge> ans = kruskal(ve, n);
    std::cout << ans[(n - 1)/2].cost << std::endl;
  }
  return 0;
}
```