#include<bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
class DisjointSet{
private:
  vector< int > rank, p;
  void link(int x,int y){
    if(rank[x] > rank[y]){
      p[y] = x;
    } else {
      p[x] = y;
      if(rank[x] == rank[y]) rank[y]++;
    }
  }
public:
  DisjointSet(int size){
    rank.resize(size,0);
    p.resize(size,0);
    for(int i = 0; i < size; i++) p[i] = i, rank[i] = 0;
  }
  void Union(int x,int y){
    if(Find(x) != Find(y)) link(Find(x),Find(y));
  }
  int Find(int x){
    return( x != p[x] ? p[x] = Find(p[x]) : p[x]);
  }
};

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

int main(){
  int n, m;
  while(cin >> n >> m, n){
    vector< edge > edges;
    for(int i = 0; i < m; i++){
      int s, t, c;
      cin >> s >> t >> c;
      edges.push_back((edge){ s - 1, t - 1, c});
    }
    sort(edges.begin(), edges.end());
    DisjointSet UnionFind(n);
    vector< int > cost;
    for(int i = 0; i < edges.size(); i++){
      edge& e = edges[i];
      if(UnionFind.Find(e.u) != UnionFind.Find(e.v)){
        UnionFind.Union( e.u, e.v);
        cost.push_back(e.cost);
      }
    }
    if(cost.size() % 2 == 0){
      cout << (cost[cost.size() / 2] + cost[cost.size() / 2 + 1]) / 2 << endl;
    } else {
      cout << cost[cost.size() / 2] << endl;
    }
  } 
}
### Prompt

Generate 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;
const int INF = 1 << 30;
class DisjointSet{
private:
  vector< int > rank, p;
  void link(int x,int y){
    if(rank[x] > rank[y]){
      p[y] = x;
    } else {
      p[x] = y;
      if(rank[x] == rank[y]) rank[y]++;
    }
  }
public:
  DisjointSet(int size){
    rank.resize(size,0);
    p.resize(size,0);
    for(int i = 0; i < size; i++) p[i] = i, rank[i] = 0;
  }
  void Union(int x,int y){
    if(Find(x) != Find(y)) link(Find(x),Find(y));
  }
  int Find(int x){
    return( x != p[x] ? p[x] = Find(p[x]) : p[x]);
  }
};

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

int main(){
  int n, m;
  while(cin >> n >> m, n){
    vector< edge > edges;
    for(int i = 0; i < m; i++){
      int s, t, c;
      cin >> s >> t >> c;
      edges.push_back((edge){ s - 1, t - 1, c});
    }
    sort(edges.begin(), edges.end());
    DisjointSet UnionFind(n);
    vector< int > cost;
    for(int i = 0; i < edges.size(); i++){
      edge& e = edges[i];
      if(UnionFind.Find(e.u) != UnionFind.Find(e.v)){
        UnionFind.Union( e.u, e.v);
        cost.push_back(e.cost);
      }
    }
    if(cost.size() % 2 == 0){
      cout << (cost[cost.size() / 2] + cost[cost.size() / 2 + 1]) / 2 << endl;
    } else {
      cout << cost[cost.size() / 2] << endl;
    }
  } 
}
```