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

constexpr int INF = 1e9;

class union_find {
public:
    union_find(int n)
        : par_(n, -1)
    {}
    void init(int n) {
        par_.assign(n, -1);
    }

    int root(int x) {
        return par_[x] < 0 ? x : par_[x] = root(par_[x]);
    }

    bool unite(int x, int y) {
        x = root(x); y = root(y);
        if(x == y) {
            return false;
        } else {
            if(par_[x] < par_[y]) { // size(x) > size(y)
                par_[x] += par_[y];
                par_[y] = x;
            } else {
                par_[y] += par_[x];
                par_[x] = y;
            }
            return true;
        }
    }

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

    int size(int x) {
        return -par_[root(x)];
    }

private:
    std::vector<int> par_;
};


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

using edges = vector<edge>;
using graph = vector<edges>;

edges kruskal (int V, edges& es) {
    sort(es.begin(), es.end());
    union_find uf(V);
    edges res;
    for(int i=0; i<es.size(); ++i) {
        edge e = es[i];
        if(!uf.same(e.s, e.t)) {
            uf.unite(e.s, e.t);
            res.push_back(e);
        }
    }
    return res;
}

int main() {
    int n, m;
    while(cin >> n >> m, n) {
        edges es(m);
        for(int i=0; i<m; ++i) {
            cin >> es[i].s >> es[i].t >> es[i].cost;
            es[i].s--;
            es[i].t--;
        }
        auto res = kruskal(n, es);
        cout << res[n/2-1].cost << endl;
    }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <set>
#include <algorithm>
using namespace std;

constexpr int INF = 1e9;

class union_find {
public:
    union_find(int n)
        : par_(n, -1)
    {}
    void init(int n) {
        par_.assign(n, -1);
    }

    int root(int x) {
        return par_[x] < 0 ? x : par_[x] = root(par_[x]);
    }

    bool unite(int x, int y) {
        x = root(x); y = root(y);
        if(x == y) {
            return false;
        } else {
            if(par_[x] < par_[y]) { // size(x) > size(y)
                par_[x] += par_[y];
                par_[y] = x;
            } else {
                par_[y] += par_[x];
                par_[x] = y;
            }
            return true;
        }
    }

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

    int size(int x) {
        return -par_[root(x)];
    }

private:
    std::vector<int> par_;
};


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

using edges = vector<edge>;
using graph = vector<edges>;

edges kruskal (int V, edges& es) {
    sort(es.begin(), es.end());
    union_find uf(V);
    edges res;
    for(int i=0; i<es.size(); ++i) {
        edge e = es[i];
        if(!uf.same(e.s, e.t)) {
            uf.unite(e.s, e.t);
            res.push_back(e);
        }
    }
    return res;
}

int main() {
    int n, m;
    while(cin >> n >> m, n) {
        edges es(m);
        for(int i=0; i<m; ++i) {
            cin >> es[i].s >> es[i].t >> es[i].cost;
            es[i].s--;
            es[i].t--;
        }
        auto res = kruskal(n, es);
        cout << res[n/2-1].cost << endl;
    }
}

```