#include <iostream>
#include <tuple>
#include <vector>
#include <algorithm>

typedef long long ll;

using namespace std;

class UnionFind {
    public:
    int n;
    vector<int> par, rank, size;  //親, 深さ, その木の要素数

    UnionFind(int n): n(n) {
        for(int i=0;i<n;++i){
            par.push_back(i);
            rank.push_back(0);
            size.push_back(1);
        }
    }

    //根を求める
    int find(int x){
        if(par[x] == x){
            return x;
        }else{
            int p = find(par[x]);
            size[x] = size[p];
            return par[x] = p;
        }
    }

    //xとyの属する集合の併合
    void unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x == y) return;
        
        size[x] += size[y];
        size[y] = size[x];
        if(rank[x] < rank[y]){
            par[x] = y;
        }else{
            par[y] = x;
            if(rank[x] == rank[y]) rank[x]++;
        }

        x = find(x);
        y = find(y);
    }

    //同じ集合に属するか否か
    bool is_same(int x, int y){
        return find(x) == find(y);
    }
};

ll solve(int n, int m) {
    vector<tuple<ll,int,int>> edges(m);
    int s,t; ll c;
    for(int i=0;i<m;++i){
        cin >> s >> t >> c;
        edges[i] = tuple<ll, int, int>{c, s, t};
    }
    sort(edges.begin(), edges.end());

    UnionFind uf(n+1);
    vector<int> edge_cnt(m);
    for(int i=0;i<m;++i){
        tie(c, s, t) = edges[i];
        if(uf.is_same(s,t)) {
            edge_cnt[i] = edge_cnt[i-1];
        } else {
            uf.unite(s, t);
            edge_cnt[i] = (i==0?0:edge_cnt[i-1]) + 1;
            if(edge_cnt[i] == ((n+1)/2)) {
                return c;
            }
        }
    }

    return 0LL;
}

int main() {
    int n, m;
    while(cin >> n >> m, n) {
        cout << solve(n, m) << endl;
    }
}

### Prompt

Your challenge is to write 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 <tuple>
#include <vector>
#include <algorithm>

typedef long long ll;

using namespace std;

class UnionFind {
    public:
    int n;
    vector<int> par, rank, size;  //親, 深さ, その木の要素数

    UnionFind(int n): n(n) {
        for(int i=0;i<n;++i){
            par.push_back(i);
            rank.push_back(0);
            size.push_back(1);
        }
    }

    //根を求める
    int find(int x){
        if(par[x] == x){
            return x;
        }else{
            int p = find(par[x]);
            size[x] = size[p];
            return par[x] = p;
        }
    }

    //xとyの属する集合の併合
    void unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x == y) return;
        
        size[x] += size[y];
        size[y] = size[x];
        if(rank[x] < rank[y]){
            par[x] = y;
        }else{
            par[y] = x;
            if(rank[x] == rank[y]) rank[x]++;
        }

        x = find(x);
        y = find(y);
    }

    //同じ集合に属するか否か
    bool is_same(int x, int y){
        return find(x) == find(y);
    }
};

ll solve(int n, int m) {
    vector<tuple<ll,int,int>> edges(m);
    int s,t; ll c;
    for(int i=0;i<m;++i){
        cin >> s >> t >> c;
        edges[i] = tuple<ll, int, int>{c, s, t};
    }
    sort(edges.begin(), edges.end());

    UnionFind uf(n+1);
    vector<int> edge_cnt(m);
    for(int i=0;i<m;++i){
        tie(c, s, t) = edges[i];
        if(uf.is_same(s,t)) {
            edge_cnt[i] = edge_cnt[i-1];
        } else {
            uf.unite(s, t);
            edge_cnt[i] = (i==0?0:edge_cnt[i-1]) + 1;
            if(edge_cnt[i] == ((n+1)/2)) {
                return c;
            }
        }
    }

    return 0LL;
}

int main() {
    int n, m;
    while(cin >> n >> m, n) {
        cout << solve(n, m) << endl;
    }
}

```