#include <bits/stdc++.h>
using namespace std;
const int MAXN=1005;
const int MAXM=10005;
int fa[MAXN];
pair< int, pair<int, int> > E[MAXM];
int find(int x){
    return fa[x]=fa[x]==x?x:find(fa[x]);
}
int main(){
    int n, m;
    while(cin>>n>>m&&(n||m)){
        for(int i=1; i<=n; i++){
            fa[i]=i;
        }
        for(int i=0, x, y, w; i<m; i++){
            scanf("%d%d%d",&x,&y,&w);
            E[i]=make_pair(w,make_pair(x,y));
        }
        sort(E,E+m);
        for(int i=0, cnt=0; i<m; i++){
            int u=find(E[i].second.first);
            int v=find(E[i].second.second);
            int w=E[i].first;
            if(u!=v){
                fa[u]=v;
                cnt++;
                if(cnt*2>=n-1){
                    printf("%d\n",w);
                    break;
                }
            }
        }
    }
    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>
using namespace std;
const int MAXN=1005;
const int MAXM=10005;
int fa[MAXN];
pair< int, pair<int, int> > E[MAXM];
int find(int x){
    return fa[x]=fa[x]==x?x:find(fa[x]);
}
int main(){
    int n, m;
    while(cin>>n>>m&&(n||m)){
        for(int i=1; i<=n; i++){
            fa[i]=i;
        }
        for(int i=0, x, y, w; i<m; i++){
            scanf("%d%d%d",&x,&y,&w);
            E[i]=make_pair(w,make_pair(x,y));
        }
        sort(E,E+m);
        for(int i=0, cnt=0; i<m; i++){
            int u=find(E[i].second.first);
            int v=find(E[i].second.second);
            int w=E[i].first;
            if(u!=v){
                fa[u]=v;
                cnt++;
                if(cnt*2>=n-1){
                    printf("%d\n",w);
                    break;
                }
            }
        }
    }
    return 0;
}
```