#include <iostream>
#include <algorithm>
#include <tuple>
#include <vector>
using namespace std;
using llong = long long;

//===
struct UnionFind {
    int n;
    vector<int> parent;

    UnionFind() {}
    UnionFind(int nmemb) {
        init(nmemb);
    };

    void init(int nmemb) {
        parent.clear();
        parent.assign(nmemb, -1);
    };

    int root(int x) {
        if (parent[x] < 0) {
            return x;
        }
        return parent[x] = root(parent[x]);
    };

    void unite(int x, int y) {
        x = root(x);
        y = root(y);

        if (x == y) return;

        if (parent[y] < parent[x]) swap(x, y);

        parent[x] += parent[y];
        parent[y] = x;
        
        return;
    };
    
    bool same(int x, int y) {
        return root(x) == root(y);
    };
    int size(int x) {
        return -(parent[root(x)]);
    };
};
//===

llong n, m;
using Edge = tuple<llong, llong, llong>;
vector<Edge> edges;

void solve() {
    edges.clear();
    UnionFind uf(n + 1);
    for (int i = 0; i < m; i++) {
        llong s, t, c;
        cin >> s >> t >> c;

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

    sort(edges.begin(), edges.end());
    llong cnt = 0;
    llong med;
    for (int i = 0; i < m; i++) {
        llong s, t, c;
        tie(c, s, t) = edges[i];

        if (uf.same(s, t)) continue;

        cnt++;
        uf.unite(s, t);

        if (cnt == n / 2) {
            med = c;
        }
    }

    cout << med << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    while (cin >> n >> m, n) {
        solve();
    }
    
    return 0;
}



### 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 <iostream>
#include <algorithm>
#include <tuple>
#include <vector>
using namespace std;
using llong = long long;

//===
struct UnionFind {
    int n;
    vector<int> parent;

    UnionFind() {}
    UnionFind(int nmemb) {
        init(nmemb);
    };

    void init(int nmemb) {
        parent.clear();
        parent.assign(nmemb, -1);
    };

    int root(int x) {
        if (parent[x] < 0) {
            return x;
        }
        return parent[x] = root(parent[x]);
    };

    void unite(int x, int y) {
        x = root(x);
        y = root(y);

        if (x == y) return;

        if (parent[y] < parent[x]) swap(x, y);

        parent[x] += parent[y];
        parent[y] = x;
        
        return;
    };
    
    bool same(int x, int y) {
        return root(x) == root(y);
    };
    int size(int x) {
        return -(parent[root(x)]);
    };
};
//===

llong n, m;
using Edge = tuple<llong, llong, llong>;
vector<Edge> edges;

void solve() {
    edges.clear();
    UnionFind uf(n + 1);
    for (int i = 0; i < m; i++) {
        llong s, t, c;
        cin >> s >> t >> c;

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

    sort(edges.begin(), edges.end());
    llong cnt = 0;
    llong med;
    for (int i = 0; i < m; i++) {
        llong s, t, c;
        tie(c, s, t) = edges[i];

        if (uf.same(s, t)) continue;

        cnt++;
        uf.unite(s, t);

        if (cnt == n / 2) {
            med = c;
        }
    }

    cout << med << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    while (cin >> n >> m, n) {
        solve();
    }
    
    return 0;
}



```