#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> P3;
typedef pair<P ,P> PP;

struct UnionFindTree {
    vector<int> par;
    vector<int> rank;
    vector<int> siz;

    void init(int n) {
        par.resize(n);
        rank.resize(n);
        siz.resize(n);
        for (int i = 0; i < n; i++) {
            par[i] = i;
            rank[i] = 0;
            siz[i] = 1;
        }
    }
    int find(int x) {
        if (par[x] == x) {
            return x;
        } else {
            return par[x] = find(par[x]);
        }
    }
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) rank[x]++;
        par[y] = x;
        siz[x] += siz[y];
    }
    bool is_same(int x, int y) {
        return find(x) == find(y);
    }
    int size(int x) {
        x = find(x);
        return siz[x];
    }
};

int n, m;
vector<P3> es;

int kruskal(){
    sort(es.begin(), es.end());
    UnionFindTree g;
    g.init(n);
    vector<int> ls;
    for(auto e : es){
        int u = e.second.first, v = e.second.second;
        if(!g.is_same(u, v)){
            g.unite(u,v);
            ls.push_back(e.first);
        }
    }
    sort(ls.begin(), ls.end());
    return ls[n/2-1];
}

int main() {
    while(cin >> n >> m, n){
        es.clear();
        for(int i=0;i<m;i++){
            int a, b, c;
            cin >> a >> b >> c;
            a--; b--;
            es.push_back({c,{a,b}});
        }
        cout << kruskal() << endl;
    }
    return 0;
}


### Prompt

Your task is to create 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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> P3;
typedef pair<P ,P> PP;

struct UnionFindTree {
    vector<int> par;
    vector<int> rank;
    vector<int> siz;

    void init(int n) {
        par.resize(n);
        rank.resize(n);
        siz.resize(n);
        for (int i = 0; i < n; i++) {
            par[i] = i;
            rank[i] = 0;
            siz[i] = 1;
        }
    }
    int find(int x) {
        if (par[x] == x) {
            return x;
        } else {
            return par[x] = find(par[x]);
        }
    }
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) rank[x]++;
        par[y] = x;
        siz[x] += siz[y];
    }
    bool is_same(int x, int y) {
        return find(x) == find(y);
    }
    int size(int x) {
        x = find(x);
        return siz[x];
    }
};

int n, m;
vector<P3> es;

int kruskal(){
    sort(es.begin(), es.end());
    UnionFindTree g;
    g.init(n);
    vector<int> ls;
    for(auto e : es){
        int u = e.second.first, v = e.second.second;
        if(!g.is_same(u, v)){
            g.unite(u,v);
            ls.push_back(e.first);
        }
    }
    sort(ls.begin(), ls.end());
    return ls[n/2-1];
}

int main() {
    while(cin >> n >> m, n){
        es.clear();
        for(int i=0;i<m;i++){
            int a, b, c;
            cin >> a >> b >> c;
            a--; b--;
            es.push_back({c,{a,b}});
        }
        cout << kruskal() << endl;
    }
    return 0;
}


```