#include <bits/stdc++.h>
#define var auto
using ll = long long;
const char newl = '\n';
template <typename T1, typename T2> inline void chmin(T1& a, T2 b) {if(a > b) a = b;}
template <typename T1, typename T2> inline void chmax(T1& a, T2 b) {if(a < b) a = b;}

using namespace std;
struct UnionFind{
    int num;
    vector<int> r,p;
    UnionFind(){}
    UnionFind(int n):num(n), r(n, 1), p(n, 0){iota(p.begin(), p.end(),0);}
    int find(int x){
        return (x==p[x] ? x : p[x] = find(p[x]));
    }
    bool same(int x, int y){
        return find(x)==find(y);
    }
    void unite(int x, int y){
        x=find(x);y=find(y);
        if (x == y) return;
        if (r[x] < r[y]) swap(x, y);
        r[x] += r[y];
        p[y]=x;
        num--;
    }
    int size(int x){
        return r[find(x)];
    }
    int count() const{
        return num;
    }
};

void solver(int n, int m) {

    struct edge {
        int s, t, c;
        edge() {}
        edge(int s, int t, int c) : s(s), t(t), c(c) {}
    };

    vector< edge > edges;
    for (int i = 0; i < m; i += 1) {
        int s, t, c;
        cin >> s >> t >> c;
        s--; t--;

        edges.emplace_back(s, t, c);
    }

    sort(edges.begin(), edges.end(), [](edge a, edge b) { return a.c < b.c; });

    vector< int > ans;
    UnionFind uf(n);

    for (auto e: edges) {
        if (uf.same(e.s, e.t)) continue;
        ans.push_back(e.c);
        uf.unite(e.s, e.t);
    }

    cout << ans[(n - 1) / 2] << endl;
    
}

signed main() {
    int n, m;
    
    while (cin >> n >> m, n) {
        solver(n, m);
    }
}




### 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 <bits/stdc++.h>
#define var auto
using ll = long long;
const char newl = '\n';
template <typename T1, typename T2> inline void chmin(T1& a, T2 b) {if(a > b) a = b;}
template <typename T1, typename T2> inline void chmax(T1& a, T2 b) {if(a < b) a = b;}

using namespace std;
struct UnionFind{
    int num;
    vector<int> r,p;
    UnionFind(){}
    UnionFind(int n):num(n), r(n, 1), p(n, 0){iota(p.begin(), p.end(),0);}
    int find(int x){
        return (x==p[x] ? x : p[x] = find(p[x]));
    }
    bool same(int x, int y){
        return find(x)==find(y);
    }
    void unite(int x, int y){
        x=find(x);y=find(y);
        if (x == y) return;
        if (r[x] < r[y]) swap(x, y);
        r[x] += r[y];
        p[y]=x;
        num--;
    }
    int size(int x){
        return r[find(x)];
    }
    int count() const{
        return num;
    }
};

void solver(int n, int m) {

    struct edge {
        int s, t, c;
        edge() {}
        edge(int s, int t, int c) : s(s), t(t), c(c) {}
    };

    vector< edge > edges;
    for (int i = 0; i < m; i += 1) {
        int s, t, c;
        cin >> s >> t >> c;
        s--; t--;

        edges.emplace_back(s, t, c);
    }

    sort(edges.begin(), edges.end(), [](edge a, edge b) { return a.c < b.c; });

    vector< int > ans;
    UnionFind uf(n);

    for (auto e: edges) {
        if (uf.same(e.s, e.t)) continue;
        ans.push_back(e.c);
        uf.unite(e.s, e.t);
    }

    cout << ans[(n - 1) / 2] << endl;
    
}

signed main() {
    int n, m;
    
    while (cin >> n >> m, n) {
        solver(n, m);
    }
}




```