#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
using namespace std;

struct edge{
    int u, v, c;
    bool operator<(const edge &e)const{
        return c < e.c;
    }
};

struct UnionFind{
    vector<int> par, sz;
    UnionFind(int n){
        par.resize(n);
        sz.resize(n, 1);
        for(int i = 0; i < n; i++) par[i] = i;
    }
    int find(int x){
        if(x == par[x]) return x;
        return par[x] = find(par[x]);
    }
    void unite(int x, int y){
        x = find(x); y = find(y);
        if(x == y) return;
        if(sz[x] < sz[y]) swap(x, y);
        sz[x] += sz[y];
        par[y] = x;
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
};

int n, m;
vector<edge> es; 

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(1){
        cin >> n >> m;
        if(n + m == 0) break;
        es.clear();
        es.resize(m);
        rep(i, 0, m){
            cin >> es[i].u >> es[i].v >> es[i].c;
            es[i].u--; es[i].v--;
        }
        UnionFind uf(n);
        sort(es.begin(), es.end());
        int ans = -1;
        int cnt = 0;
        rep(i, 0, m){
            if(!uf.same(es[i].u, es[i].v)){
                uf.unite(es[i].u, es[i].v);
                cnt++;
                if(cnt == n / 2){
                    ans = es[i].c;
                }
            }
        }
        cout << ans << endl;
    }
}

### Prompt

Please formulate 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 <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
using namespace std;

struct edge{
    int u, v, c;
    bool operator<(const edge &e)const{
        return c < e.c;
    }
};

struct UnionFind{
    vector<int> par, sz;
    UnionFind(int n){
        par.resize(n);
        sz.resize(n, 1);
        for(int i = 0; i < n; i++) par[i] = i;
    }
    int find(int x){
        if(x == par[x]) return x;
        return par[x] = find(par[x]);
    }
    void unite(int x, int y){
        x = find(x); y = find(y);
        if(x == y) return;
        if(sz[x] < sz[y]) swap(x, y);
        sz[x] += sz[y];
        par[y] = x;
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
};

int n, m;
vector<edge> es; 

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(1){
        cin >> n >> m;
        if(n + m == 0) break;
        es.clear();
        es.resize(m);
        rep(i, 0, m){
            cin >> es[i].u >> es[i].v >> es[i].c;
            es[i].u--; es[i].v--;
        }
        UnionFind uf(n);
        sort(es.begin(), es.end());
        int ans = -1;
        int cnt = 0;
        rep(i, 0, m){
            if(!uf.same(es[i].u, es[i].v)){
                uf.unite(es[i].u, es[i].v);
                cnt++;
                if(cnt == n / 2){
                    ans = es[i].c;
                }
            }
        }
        cout << ans << endl;
    }
}

```