#include <bits/stdc++.h>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;

struct Edge{
    int s, t, c;
    bool operator < (const Edge& e) const {
        if(c != e.c) return c < e.c;
        return make_pair(s, t) < make_pair(e.s, e.t);
    }
};

struct UnionFind{
    vector<int> data;
    UnionFind(int N) : data(N) {
        REP(i, N) data[i] = i;
    }
    bool merge(int x, int y){
        x = root(x);
        y = root(y);
        if(x == y) return false;
        data[x] = y;
        return true;
    }
    int root(int x){
        if(x == data[x]) return x;
        return data[x] = root(data[x]);
    }
};

int main(){
    int N, M;
    while(cin >> N >> M && N){
        vector<Edge> edges;
        REP(i, M){
            Edge e;
            cin >> e.s >> e.t >> e.c;
            e.s--; e.t--;
            edges.push_back(e);
        }
        sort(edges.begin(), edges.end());
        UnionFind uf(N);
        vector<int> costs;
        REP(i, M){
            Edge& e = edges[i];
            if(uf.merge(e.s, e.t)){
                costs.push_back(e.c);
            }
        }
        cout << costs[ costs.size() / 2 ] << endl;
    }
    return 0;
}
### Prompt

Your challenge is to write 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>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;

struct Edge{
    int s, t, c;
    bool operator < (const Edge& e) const {
        if(c != e.c) return c < e.c;
        return make_pair(s, t) < make_pair(e.s, e.t);
    }
};

struct UnionFind{
    vector<int> data;
    UnionFind(int N) : data(N) {
        REP(i, N) data[i] = i;
    }
    bool merge(int x, int y){
        x = root(x);
        y = root(y);
        if(x == y) return false;
        data[x] = y;
        return true;
    }
    int root(int x){
        if(x == data[x]) return x;
        return data[x] = root(data[x]);
    }
};

int main(){
    int N, M;
    while(cin >> N >> M && N){
        vector<Edge> edges;
        REP(i, M){
            Edge e;
            cin >> e.s >> e.t >> e.c;
            e.s--; e.t--;
            edges.push_back(e);
        }
        sort(edges.begin(), edges.end());
        UnionFind uf(N);
        vector<int> costs;
        REP(i, M){
            Edge& e = edges[i];
            if(uf.merge(e.s, e.t)){
                costs.push_back(e.c);
            }
        }
        cout << costs[ costs.size() / 2 ] << endl;
    }
    return 0;
}
```