#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <functional>
#include <utility>

using namespace std;

typedef pair<int, int> pii;

struct UF {
    vector<int> data;

    UF(int n) {
        data.resize(n, -1);
    }

    int root(int i) {
        return data[i] < 0 ? i : data[i] = root(data[i]);
    }

    int size(int i) {
        return -data[root(i)];
    }

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

        if (x == y) return ;

        if (size(x) < size(y)) swap(x, y);

        data[x] += data[y];
        data[y] = x;
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }
};

int N, M;

int main()
{
    while (cin >> N >> M, N) {
        UF uf(N);

        vector<pair<int, pii>> edges;

        for (int i = 0; i < M; ++i) {
            int s, t, c; cin >> s >> t >> c;

            --s, --t;

            edges.push_back(make_pair(c, pii(s, t)));
        }

        sort(begin(edges), end(edges));

        vector<int> a;

        for (pair<int, pii> e : edges) {
            int cost = e.first;
            pii info = e.second;
            if (!uf.same(info.first, info.second)) {
                uf.unite(info.first, info.second);
                a.push_back(cost);
            }
        }

        sort(begin(a), end(a));

        cout << a[a.size() / 2] << 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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <functional>
#include <utility>

using namespace std;

typedef pair<int, int> pii;

struct UF {
    vector<int> data;

    UF(int n) {
        data.resize(n, -1);
    }

    int root(int i) {
        return data[i] < 0 ? i : data[i] = root(data[i]);
    }

    int size(int i) {
        return -data[root(i)];
    }

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

        if (x == y) return ;

        if (size(x) < size(y)) swap(x, y);

        data[x] += data[y];
        data[y] = x;
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }
};

int N, M;

int main()
{
    while (cin >> N >> M, N) {
        UF uf(N);

        vector<pair<int, pii>> edges;

        for (int i = 0; i < M; ++i) {
            int s, t, c; cin >> s >> t >> c;

            --s, --t;

            edges.push_back(make_pair(c, pii(s, t)));
        }

        sort(begin(edges), end(edges));

        vector<int> a;

        for (pair<int, pii> e : edges) {
            int cost = e.first;
            pii info = e.second;
            if (!uf.same(info.first, info.second)) {
                uf.unite(info.first, info.second);
                a.push_back(cost);
            }
        }

        sort(begin(a), end(a));

        cout << a[a.size() / 2] << endl;
    }
}
```