#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int par[1000];

void init(int n){
    REP(i,n){
        par[i] = i;
    }
}

int find(int x){
    if (par[x] == x){
        return x;
    }else{
        return par[x] = find(par[x]);
    }
}

bool same(int x, int y){
    return find(x) == find(y);
}

void merge(int x, int y){
    x = find(x);
    y = find(y);
    if (x==y) return;
    par[x] = y;
}

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

vector<edge> es;
int n;

bool comp(const edge &e1, const edge &e2){
    return e1.cost < e2.cost;
}

int kruskal(){
    sort(es.begin(), es.end(), comp);
    init(n);
    VI e_used;
    REP(i,es.size()){
        edge e = es[i];
        if (!same(e.u, e.v)){
            merge(e.u, e.v);
            e_used.push_back(e.cost);
        }
    }
    return e_used[n/2-1];
}

int main() {
    int m;
    while (cin >> n >> m && n){
        es.clear();
        while (m--){
            edge e;
            cin >> e.u >> e.v >> e.cost;
            e.u--;
            e.v--;
            es.push_back(e);
        }
        cout << kruskal() << endl;
    }

    return 0;
}
### Prompt

In CPP, your task is to solve 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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int par[1000];

void init(int n){
    REP(i,n){
        par[i] = i;
    }
}

int find(int x){
    if (par[x] == x){
        return x;
    }else{
        return par[x] = find(par[x]);
    }
}

bool same(int x, int y){
    return find(x) == find(y);
}

void merge(int x, int y){
    x = find(x);
    y = find(y);
    if (x==y) return;
    par[x] = y;
}

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

vector<edge> es;
int n;

bool comp(const edge &e1, const edge &e2){
    return e1.cost < e2.cost;
}

int kruskal(){
    sort(es.begin(), es.end(), comp);
    init(n);
    VI e_used;
    REP(i,es.size()){
        edge e = es[i];
        if (!same(e.u, e.v)){
            merge(e.u, e.v);
            e_used.push_back(e.cost);
        }
    }
    return e_used[n/2-1];
}

int main() {
    int m;
    while (cin >> n >> m && n){
        es.clear();
        while (m--){
            edge e;
            cin >> e.u >> e.v >> e.cost;
            e.u--;
            e.v--;
            es.push_back(e);
        }
        cout << kruskal() << endl;
    }

    return 0;
}
```