#include "bits/stdc++.h"

using namespace std;

struct Edge{
    int from,to,cost;
    Edge(int from,int to,int cost): from(from), to(to), cost(cost){}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

bool operator < (const Edge &e, const Edge &f){
    return e.cost < f.cost;
}

struct UF {
	vector<int> data;
	UF(int size) : data(size, -1) { }
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool find(int x, int y) {
		return root(x) == root(y);
	}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {
		return -data[root(x)];
	}
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n,m;
    while(cin>>n>>m,n){
        UF uf(n);
        Edges es;
        vector<int> ans;
        for(int i=0;i<m;i++){
            int st,gt,c;cin>>st>>gt>>c;
            st--;gt--;
            es.push_back(Edge(st,gt,c));
        }
        sort(es.begin(),es.end());
        for(int i=0;i<m;i++){
            int st=es[i].from,gt=es[i].to;
            if(!uf.find(st,gt)){
                uf.unite(st,gt);
                ans.push_back(es[i].cost);
            }
        }
        cout<<ans[(n-1)/2]<<endl;
    }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

struct Edge{
    int from,to,cost;
    Edge(int from,int to,int cost): from(from), to(to), cost(cost){}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

bool operator < (const Edge &e, const Edge &f){
    return e.cost < f.cost;
}

struct UF {
	vector<int> data;
	UF(int size) : data(size, -1) { }
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool find(int x, int y) {
		return root(x) == root(y);
	}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {
		return -data[root(x)];
	}
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n,m;
    while(cin>>n>>m,n){
        UF uf(n);
        Edges es;
        vector<int> ans;
        for(int i=0;i<m;i++){
            int st,gt,c;cin>>st>>gt>>c;
            st--;gt--;
            es.push_back(Edge(st,gt,c));
        }
        sort(es.begin(),es.end());
        for(int i=0;i<m;i++){
            int st=es[i].from,gt=es[i].to;
            if(!uf.find(st,gt)){
                uf.unite(st,gt);
                ans.push_back(es[i].cost);
            }
        }
        cout<<ans[(n-1)/2]<<endl;
    }
}

```