#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <tuple>

using namespace std;

struct UnionFind {
    vector<int> par, sz;

    UnionFind(int N) : par(N), sz(N, 1) {
        iota(par.begin(), par.end(), 0);
    }

    int find(int v) {
        return par[v] == v ? v : par[v] = find(par[v]);
    }

    void unite(int u, int v) {
        u = find(u);
        v = find(v);
        if (sz[u] < sz[v]) swap(u, v);

        sz[u] += sz[v];
        par[v] = u;
    }
};

bool solve() {
    int N, M;
    cin >> N >> M;
    if (N == 0) return false;

    vector<tuple<int, int, int>> edges(M);
    for (auto& e : edges) {
        int u, v, c;
        cin >> u >> v >> c;
        --u, --v;
        e = make_tuple(c, u, v);
    }

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

    UnionFind uf(N);
    int cnt = 0;
    for (auto e : edges) {
        int u, v, c;
        tie(c, u, v) = e;
        if (uf.find(u) == uf.find(v)) continue;
        if (++cnt == N / 2) {
            cout << c << endl;
            break;
        }
        uf.unite(u, v);
    }
    return true;
}

int main() {
    while (solve()) {}
    return 0;
}


### Prompt

Please provide a cpp coded solution 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 <vector>
#include <numeric>
#include <tuple>

using namespace std;

struct UnionFind {
    vector<int> par, sz;

    UnionFind(int N) : par(N), sz(N, 1) {
        iota(par.begin(), par.end(), 0);
    }

    int find(int v) {
        return par[v] == v ? v : par[v] = find(par[v]);
    }

    void unite(int u, int v) {
        u = find(u);
        v = find(v);
        if (sz[u] < sz[v]) swap(u, v);

        sz[u] += sz[v];
        par[v] = u;
    }
};

bool solve() {
    int N, M;
    cin >> N >> M;
    if (N == 0) return false;

    vector<tuple<int, int, int>> edges(M);
    for (auto& e : edges) {
        int u, v, c;
        cin >> u >> v >> c;
        --u, --v;
        e = make_tuple(c, u, v);
    }

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

    UnionFind uf(N);
    int cnt = 0;
    for (auto e : edges) {
        int u, v, c;
        tie(c, u, v) = e;
        if (uf.find(u) == uf.find(v)) continue;
        if (++cnt == N / 2) {
            cout << c << endl;
            break;
        }
        uf.unite(u, v);
    }
    return true;
}

int main() {
    while (solve()) {}
    return 0;
}


```