#include <bits/stdc++.h>
using namespace std;

typedef struct Edge{
    int w, s, t;
    
    bool operator < (const Edge r)const{
        return w < r.w;
    };
}Edge;

class UnionFind {
    vector<int> par;
    int cnt;
public:
    UnionFind(int size_) : par(size_, -1), cnt(size_) { }
    void unite(int x, int y){
        if ((x = find(x)) != (y = find(y))) {
            if (par[y] < par[x]) swap(x, y);
            par[x] += par[y]; par[y] = x; cnt--;
        }
    }
    bool same(int x, int y){ return find(x) == find(y); }
    int find(int x){ return par[x] < 0 ? x : par[x] = find(par[x]); }
    int size(int x){ return -par[find(x)]; }
    int size(){ return cnt; }
};

int N, M;

int main(){
    while(cin >> N >> M, N|M){
        auto uf = UnionFind(N);
        
        vector<Edge> edges;
        for(int i=0; i<M; i++){
            int s, t, w;
            cin >> s >> t >> w; s--; t--;
            edges.push_back({w, s, t});
        }
        sort(edges.begin(), edges.end());
        
        int cnt = 0;
        for(auto &e: edges){
            if(!uf.same(e.s, e.t)){
                uf.unite(e.s, e.t);
                cnt++;
            }
            if(cnt == N/2){
                cout << e.w << endl;
                break;
            }
        }
    }
    
    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 <bits/stdc++.h>
using namespace std;

typedef struct Edge{
    int w, s, t;
    
    bool operator < (const Edge r)const{
        return w < r.w;
    };
}Edge;

class UnionFind {
    vector<int> par;
    int cnt;
public:
    UnionFind(int size_) : par(size_, -1), cnt(size_) { }
    void unite(int x, int y){
        if ((x = find(x)) != (y = find(y))) {
            if (par[y] < par[x]) swap(x, y);
            par[x] += par[y]; par[y] = x; cnt--;
        }
    }
    bool same(int x, int y){ return find(x) == find(y); }
    int find(int x){ return par[x] < 0 ? x : par[x] = find(par[x]); }
    int size(int x){ return -par[find(x)]; }
    int size(){ return cnt; }
};

int N, M;

int main(){
    while(cin >> N >> M, N|M){
        auto uf = UnionFind(N);
        
        vector<Edge> edges;
        for(int i=0; i<M; i++){
            int s, t, w;
            cin >> s >> t >> w; s--; t--;
            edges.push_back({w, s, t});
        }
        sort(edges.begin(), edges.end());
        
        int cnt = 0;
        for(auto &e: edges){
            if(!uf.same(e.s, e.t)){
                uf.unite(e.s, e.t);
                cnt++;
            }
            if(cnt == N/2){
                cout << e.w << endl;
                break;
            }
        }
    }
    
    return 0;
}
```