#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class unionfind{
public:
    vector<int> v;
    vector<int> rank;

    unionfind(int n){
        v = vector<int>(n);
        rank = vector<int>(n, 0);
        for(int i=0; i<n; i++) v[i]=i;
    }
    int find(int x){
        if(v[x] != x){
            v[x] = find(v[x]);
        }
        return v[x];
    }
    void marge(int a, int b){
        a = find(a);
        b = find(b);
        if(rank[a] < rank[b]){
            v[a] = b;
        }else{
            v[b] = a;
            if(rank[a] == rank[b]){
                rank[a]++;
            }
        }
    }
};

struct edge{
    int s,t;
    int cost;
    edge(int s, int t, int c):s(s), t(t), cost(c){}
    edge(){}
    bool operator < (const edge &a) const{
        return cost < a.cost;
    }
};

int main(){
    while(1){
        int n,m;
        cin >> n >> m;
        if(n==0) break;

        vector<edge> ve(m);
        for(int i=0; i<m; i++){
            int s,t,c;
            cin >> s >> t >> c;
            s--; t--;
            ve[i] = edge(s,t,c);
        }
        sort(ve.begin(), ve.end());

        unionfind uf(n);
        int count = 0;
        for(int i=0; i<m; i++){
            if(uf.find(ve[i].s) != uf.find(ve[i].t)){
                uf.marge(ve[i].s, ve[i].t);
                count++;
                if(count == n/2){
                    cout << ve[i].cost << endl;
                    break;
                }
            }
        }
    }
    return 0;
}

### Prompt

Please formulate 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class unionfind{
public:
    vector<int> v;
    vector<int> rank;

    unionfind(int n){
        v = vector<int>(n);
        rank = vector<int>(n, 0);
        for(int i=0; i<n; i++) v[i]=i;
    }
    int find(int x){
        if(v[x] != x){
            v[x] = find(v[x]);
        }
        return v[x];
    }
    void marge(int a, int b){
        a = find(a);
        b = find(b);
        if(rank[a] < rank[b]){
            v[a] = b;
        }else{
            v[b] = a;
            if(rank[a] == rank[b]){
                rank[a]++;
            }
        }
    }
};

struct edge{
    int s,t;
    int cost;
    edge(int s, int t, int c):s(s), t(t), cost(c){}
    edge(){}
    bool operator < (const edge &a) const{
        return cost < a.cost;
    }
};

int main(){
    while(1){
        int n,m;
        cin >> n >> m;
        if(n==0) break;

        vector<edge> ve(m);
        for(int i=0; i<m; i++){
            int s,t,c;
            cin >> s >> t >> c;
            s--; t--;
            ve[i] = edge(s,t,c);
        }
        sort(ve.begin(), ve.end());

        unionfind uf(n);
        int count = 0;
        for(int i=0; i<m; i++){
            if(uf.find(ve[i].s) != uf.find(ve[i].t)){
                uf.marge(ve[i].s, ve[i].t);
                count++;
                if(count == n/2){
                    cout << ve[i].cost << endl;
                    break;
                }
            }
        }
    }
    return 0;
}

```