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

struct UF{
    vector<int> p;
    int n;

    UF(int siz){
        n = siz;
        p.resize(n, 0);
        for(int i = 0; i < n; i++)  p[i] = i;
    }

    int parent(int x){
        if(p[x] != x)   p[x] = parent(p[x]);
        return p[x];
    }

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

    void unite(int x, int y){
        x = parent(x), y = parent(y);
        p[x] = y;
    }
};

// MSTやるだけ、は流石に無い？
int main(){
    int n, m;
    while(cin >> n >> m, n+m){
        priority_queue<pair<int,pair<int,int>>> pq;
        int s, t, c;
        for(int i = 0; i < m; i++){
            cin >> s >> t >> c;
            s--, t--;
            pq.push({-c, {s,t}});
        }
        UF uf(n);
        int cnt = 0;
        while(true){
            auto p = pq.top();    pq.pop();
            int i = p.second.first, j = p.second.second;
            if(!uf.same(i,j)){
                uf.unite(i,j);
                cnt++;
                if(cnt == n/2){
                    cout << -p.first << 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
#include<iostream>
#include<vector>
#include<queue>
using namespace std;

struct UF{
    vector<int> p;
    int n;

    UF(int siz){
        n = siz;
        p.resize(n, 0);
        for(int i = 0; i < n; i++)  p[i] = i;
    }

    int parent(int x){
        if(p[x] != x)   p[x] = parent(p[x]);
        return p[x];
    }

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

    void unite(int x, int y){
        x = parent(x), y = parent(y);
        p[x] = y;
    }
};

// MSTやるだけ、は流石に無い？
int main(){
    int n, m;
    while(cin >> n >> m, n+m){
        priority_queue<pair<int,pair<int,int>>> pq;
        int s, t, c;
        for(int i = 0; i < m; i++){
            cin >> s >> t >> c;
            s--, t--;
            pq.push({-c, {s,t}});
        }
        UF uf(n);
        int cnt = 0;
        while(true){
            auto p = pq.top();    pq.pop();
            int i = p.second.first, j = p.second.second;
            if(!uf.same(i,j)){
                uf.unite(i,j);
                cnt++;
                if(cnt == n/2){
                    cout << -p.first << endl;
                    break;
                }
            }
        }
    }
    return 0;
}

```