#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

int N,M;

struct Node{
	bool operator<(const struct Node &arg) const{
		return weight < arg.weight;
	};
	int from,to,weight;
};

int parent[1001],height[1001];
Node* nodes;

int getParent(int id){
	if(parent[id] == id)return id;
	else{
		return parent[id] = getParent(parent[id]);
	}
}

bool isSame(int a, int b){
	return getParent(a) == getParent(b);
}

void unite(int a,int b){
	int a_parent = getParent(a);
	int b_parent = getParent(b);

	if(a_parent == b_parent)return;

	if(height[a_parent] > height[b_parent]){
		parent[b_parent] = a_parent;
	}else{
		if(height[a_parent] == height[b_parent])height[b_parent]++;
		parent[a_parent] = b_parent;
	}
}

void init(){
	for(int i = 1; i <= N; i++){
		parent[i] = i;
		height[i] = 0;
	}
}

void func(){

	for(int i = 0; i < M; i++){
		scanf("%d %d %d",&nodes[i].from,&nodes[i].to,&nodes[i].weight);
	}

	sort(nodes,nodes+M);

	int count = 0,ans;

	init();

	for(int i = 0; i < M; i++){
		if(!isSame(nodes[i].from,nodes[i].to)){
			unite(nodes[i].from,nodes[i].to);
			if(count == (N-1)/2){
				ans = nodes[i].weight;
				break;
			}
			count++;
		}
	}

	printf("%d\n",ans);
}

int main(){

	nodes = (Node*)malloc(sizeof(Node)*10000);

	while(true){
		scanf("%d %d",&N,&M);
		if(N == 0 && M == 0)break;

		func();
	}
}
### 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 <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

int N,M;

struct Node{
	bool operator<(const struct Node &arg) const{
		return weight < arg.weight;
	};
	int from,to,weight;
};

int parent[1001],height[1001];
Node* nodes;

int getParent(int id){
	if(parent[id] == id)return id;
	else{
		return parent[id] = getParent(parent[id]);
	}
}

bool isSame(int a, int b){
	return getParent(a) == getParent(b);
}

void unite(int a,int b){
	int a_parent = getParent(a);
	int b_parent = getParent(b);

	if(a_parent == b_parent)return;

	if(height[a_parent] > height[b_parent]){
		parent[b_parent] = a_parent;
	}else{
		if(height[a_parent] == height[b_parent])height[b_parent]++;
		parent[a_parent] = b_parent;
	}
}

void init(){
	for(int i = 1; i <= N; i++){
		parent[i] = i;
		height[i] = 0;
	}
}

void func(){

	for(int i = 0; i < M; i++){
		scanf("%d %d %d",&nodes[i].from,&nodes[i].to,&nodes[i].weight);
	}

	sort(nodes,nodes+M);

	int count = 0,ans;

	init();

	for(int i = 0; i < M; i++){
		if(!isSame(nodes[i].from,nodes[i].to)){
			unite(nodes[i].from,nodes[i].to);
			if(count == (N-1)/2){
				ans = nodes[i].weight;
				break;
			}
			count++;
		}
	}

	printf("%d\n",ans);
}

int main(){

	nodes = (Node*)malloc(sizeof(Node)*10000);

	while(true){
		scanf("%d %d",&N,&M);
		if(N == 0 && M == 0)break;

		func();
	}
}
```