#include <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;


class UnionFind {
    vector<int> uni;
    int n;
public:
    explicit UnionFind(int n) : uni(static_cast<u32>(n), -1) , n(n){};

    int root(int a){
        if (uni[a] < 0) return a;
        else return (uni[a] = root(uni[a]));
    }

    bool unite(int a, int b) {
        a = root(a);
        b = root(b);
        if(a == b) return false;
        if(uni[a] > uni[b]) swap(a, b);
        uni[a] += uni[b];
        uni[b] = a;
        return true;
    }
};


int main() {
    int n, m;
    while(cin >> n >> m, n){
        vector<tuple<int, int, int>> v(m);
        for (int i = 0; i < m; ++i) {
            int s, t, c;
            cin >> s >> t >> c;
            s--; t--;
            v[i] = make_tuple(c, s, t);
        }
        sort(v.begin(),v.end());
        vector<int> e;
        UnionFind uf(n);
        for(auto &&i : v){
            int c, s, t; tie(c, s, t) = i;
            if(uf.unite(s, t)) e.emplace_back(c);
        }
        cout << e[(n-1)/2] << "\n";
    }
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;


class UnionFind {
    vector<int> uni;
    int n;
public:
    explicit UnionFind(int n) : uni(static_cast<u32>(n), -1) , n(n){};

    int root(int a){
        if (uni[a] < 0) return a;
        else return (uni[a] = root(uni[a]));
    }

    bool unite(int a, int b) {
        a = root(a);
        b = root(b);
        if(a == b) return false;
        if(uni[a] > uni[b]) swap(a, b);
        uni[a] += uni[b];
        uni[b] = a;
        return true;
    }
};


int main() {
    int n, m;
    while(cin >> n >> m, n){
        vector<tuple<int, int, int>> v(m);
        for (int i = 0; i < m; ++i) {
            int s, t, c;
            cin >> s >> t >> c;
            s--; t--;
            v[i] = make_tuple(c, s, t);
        }
        sort(v.begin(),v.end());
        vector<int> e;
        UnionFind uf(n);
        for(auto &&i : v){
            int c, s, t; tie(c, s, t) = i;
            if(uf.unite(s, t)) e.emplace_back(c);
        }
        cout << e[(n-1)/2] << "\n";
    }
    return 0;
}

```