#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;
typedef pair<int,P> PP;

class Union_Find
{
public:
    vector<int> par;

    void init(int n) {
        par.resize(n);
        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) par[y] = x;
        else par[x] = y;
    }
};

int N, M;

void solve() {
    vector<PP> V;
    for (int i = 0; i < M; i++) {
        int x, y, c;
        cin >> x >> y >> c;
        x--; y--;
        V.push_back(PP(c, P(x, y)));
    }

    sort(V.begin(), V.end());

    vector<bool> used(N, false);
    int count = 0;
    Union_Find uf;
    uf.init(N);
    for (int i = 0; i < M; i++) {
        PP pp = V[i];
        int cost = pp.first;
        int x = pp.second.first;
        int y = pp.second.second;
        if (uf.find(x) == uf.find(y)) continue;
        count++;
        uf.unite(x, y);

        if (count == (N + 1) / 2) {
            cout << cost << endl;
            return ;
        }
    }
}

int main() {
    while (cin >> N >> M, N || M) solve();

    return 0;
}
### Prompt

Generate 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 pair<int,int> P;
typedef pair<int,P> PP;

class Union_Find
{
public:
    vector<int> par;

    void init(int n) {
        par.resize(n);
        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) par[y] = x;
        else par[x] = y;
    }
};

int N, M;

void solve() {
    vector<PP> V;
    for (int i = 0; i < M; i++) {
        int x, y, c;
        cin >> x >> y >> c;
        x--; y--;
        V.push_back(PP(c, P(x, y)));
    }

    sort(V.begin(), V.end());

    vector<bool> used(N, false);
    int count = 0;
    Union_Find uf;
    uf.init(N);
    for (int i = 0; i < M; i++) {
        PP pp = V[i];
        int cost = pp.first;
        int x = pp.second.first;
        int y = pp.second.second;
        if (uf.find(x) == uf.find(y)) continue;
        count++;
        uf.unite(x, y);

        if (count == (N + 1) / 2) {
            cout << cost << endl;
            return ;
        }
    }
}

int main() {
    while (cin >> N >> M, N || M) solve();

    return 0;
}
```