#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct DisjointSet/*{{{*/
{
    vector<int> parent;

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

    explicit DisjointSet(int n) : parent(n, -1) {}

    bool unite(int x, int y)
    {
        const int a = root(x);
        const int b = root(y);
        if (a != b) {
            if (parent[a] < parent[b]) {
                parent[a] += parent[b];
                parent[b] = a;
            } else {
                parent[b] += parent[a];
                parent[a] = b;
            }
            return true;
        } else {
            return false;
        }
    }

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

struct Edge {
    int a, b;
    int cost;

    bool operator <(const Edge &other) const {
        return cost < other.cost;
    }
};

bool solve() {
    int N, M;
    if(!(cin >> N >> M)) return false;
    if(!N && !M) return false;

    vector<Edge> edges(M);
    for(int i = 0; i < M; ++i) {
        Edge &e = edges[i];
        cin >> e.a >> e.b >> e.cost;
        e.a--;
        e.b--;
    }
    sort(edges.begin(), edges.end());
    DisjointSet ds(N);
    int cnt = 0;
    for(Edge &e : edges) {
        if(ds.unite(e.a, e.b)) {
            ++cnt;
            if(cnt == N/2) {
                cout << e.cost << endl;
                break;
            }
        }
    }
    return true;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    while(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 <vector>
#include <algorithm>

using namespace std;

struct DisjointSet/*{{{*/
{
    vector<int> parent;

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

    explicit DisjointSet(int n) : parent(n, -1) {}

    bool unite(int x, int y)
    {
        const int a = root(x);
        const int b = root(y);
        if (a != b) {
            if (parent[a] < parent[b]) {
                parent[a] += parent[b];
                parent[b] = a;
            } else {
                parent[b] += parent[a];
                parent[a] = b;
            }
            return true;
        } else {
            return false;
        }
    }

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

struct Edge {
    int a, b;
    int cost;

    bool operator <(const Edge &other) const {
        return cost < other.cost;
    }
};

bool solve() {
    int N, M;
    if(!(cin >> N >> M)) return false;
    if(!N && !M) return false;

    vector<Edge> edges(M);
    for(int i = 0; i < M; ++i) {
        Edge &e = edges[i];
        cin >> e.a >> e.b >> e.cost;
        e.a--;
        e.b--;
    }
    sort(edges.begin(), edges.end());
    DisjointSet ds(N);
    int cnt = 0;
    for(Edge &e : edges) {
        if(ds.unite(e.a, e.b)) {
            ++cnt;
            if(cnt == N/2) {
                cout << e.cost << endl;
                break;
            }
        }
    }
    return true;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    while(solve()) ;
    return 0;
}
```