#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;

#define MAX_N 200000

int par[MAX_N];
int rnk[MAX_N];

// 初期化
void init(int n){
	for(int i = 0; i < n; i++){
		par[i] = i;
		rnk[i] = 0;
	}
}

// x の親を返す
int find(int x){
	if(par[x] == x){
		return x;
	}
	return par[x] = find(par[x]);
}

// x と y を併合する
void unite(int x, int y){
	int x_par = find(x);
	int y_par = find(y);
	if(x_par == y_par){
		return;
	}
	
	if(rnk[x_par] < rnk[y_par]){
		par[x_par] = y_par;
	}else{
		par[y_par] = x_par;
		if(rnk[x_par] == rnk[y_par]){
			rnk[x_par]++;
		}
	}
}

// x, y が同じグループか判定
bool same(int x, int y){
	return (find(x) == find(y));
}

struct edge{
	int u, v, cost;
};

bool cmp(edge e1, edge e2){
	return e1.cost < e2.cost;
}

int main(){
	
	while(true){
		int n, m;
		cin >> n >> m;
		if(n == 0){
			break;
		}
		vector<edge> v;
		for(int i = 0; i < m; i++){
			edge e;
			cin >> e.u >> e.v >> e.cost;
			v.push_back(e);
		}
		sort(v.begin(), v.end(), cmp);
		init(2000);
		
		int cnt = 0;
		for(int i = 0; i < m; i++){
			edge e = v[i];
			if(!same(e.u, e.v)){
				unite(e.u, e.v);
				cnt++;
			}
			if(cnt == n / 2){
				cout << e.cost << endl;
				break;
			}
		}
	}
	
	return 0;
}



### Prompt

Create a solution in Cpp for 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
#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;

#define MAX_N 200000

int par[MAX_N];
int rnk[MAX_N];

// 初期化
void init(int n){
	for(int i = 0; i < n; i++){
		par[i] = i;
		rnk[i] = 0;
	}
}

// x の親を返す
int find(int x){
	if(par[x] == x){
		return x;
	}
	return par[x] = find(par[x]);
}

// x と y を併合する
void unite(int x, int y){
	int x_par = find(x);
	int y_par = find(y);
	if(x_par == y_par){
		return;
	}
	
	if(rnk[x_par] < rnk[y_par]){
		par[x_par] = y_par;
	}else{
		par[y_par] = x_par;
		if(rnk[x_par] == rnk[y_par]){
			rnk[x_par]++;
		}
	}
}

// x, y が同じグループか判定
bool same(int x, int y){
	return (find(x) == find(y));
}

struct edge{
	int u, v, cost;
};

bool cmp(edge e1, edge e2){
	return e1.cost < e2.cost;
}

int main(){
	
	while(true){
		int n, m;
		cin >> n >> m;
		if(n == 0){
			break;
		}
		vector<edge> v;
		for(int i = 0; i < m; i++){
			edge e;
			cin >> e.u >> e.v >> e.cost;
			v.push_back(e);
		}
		sort(v.begin(), v.end(), cmp);
		init(2000);
		
		int cnt = 0;
		for(int i = 0; i < m; i++){
			edge e = v[i];
			if(!same(e.u, e.v)){
				unite(e.u, e.v);
				cnt++;
			}
			if(cnt == n / 2){
				cout << e.cost << endl;
				break;
			}
		}
	}
	
	return 0;
}



```