#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

struct Edge {
    int from, to, cost;
    Edge(int a, int b, int c) : from(a), to(b), cost(c) {}
    bool operator<(const Edge &x) const {
        return cost < x.cost;
    }
};

struct UnionFind {
    int node[1010];
    UnionFind() {
        memset(node, -1, sizeof(node));
    }
    int find(int x) {
        return node[x] < 0 ? x : node[x] = find(node[x]);
    }
    int size(int x) {
        return -node[find(x)];
    }
    bool unite(int x, int y) {
        x = find(x), y = find(y);
        if(x == y) return false;
        if(size(x) > size(y)) swap(x, y);
        node[y] += node[x];
        node[x] = y;
        return true;
    }
};

int main() {
    int N, M;
    while(1) {
        scanf("%d%d", &N, &M);
        if(!N) break;
        vector<Edge> edges;
        for(int i=0; i<M; i++) {
            int s, t, c;
            scanf("%d%d%d", &s, &t, &c); s--; t--;
            edges.push_back(Edge(s, t, c));
        }

        sort(edges.begin(), edges.end());
        vector<int> use_cost;
        UnionFind uf;
        for(int i=0; i<M; i++) {
            Edge e = edges[i];
            if(uf.unite(e.from, e.to)) {
                use_cost.push_back(e.cost);
            }
        }
        int mid = (N-1) / 2;
        printf("%d\n", use_cost[mid]);
    }
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

struct Edge {
    int from, to, cost;
    Edge(int a, int b, int c) : from(a), to(b), cost(c) {}
    bool operator<(const Edge &x) const {
        return cost < x.cost;
    }
};

struct UnionFind {
    int node[1010];
    UnionFind() {
        memset(node, -1, sizeof(node));
    }
    int find(int x) {
        return node[x] < 0 ? x : node[x] = find(node[x]);
    }
    int size(int x) {
        return -node[find(x)];
    }
    bool unite(int x, int y) {
        x = find(x), y = find(y);
        if(x == y) return false;
        if(size(x) > size(y)) swap(x, y);
        node[y] += node[x];
        node[x] = y;
        return true;
    }
};

int main() {
    int N, M;
    while(1) {
        scanf("%d%d", &N, &M);
        if(!N) break;
        vector<Edge> edges;
        for(int i=0; i<M; i++) {
            int s, t, c;
            scanf("%d%d%d", &s, &t, &c); s--; t--;
            edges.push_back(Edge(s, t, c));
        }

        sort(edges.begin(), edges.end());
        vector<int> use_cost;
        UnionFind uf;
        for(int i=0; i<M; i++) {
            Edge e = edges[i];
            if(uf.unite(e.from, e.to)) {
                use_cost.push_back(e.cost);
            }
        }
        int mid = (N-1) / 2;
        printf("%d\n", use_cost[mid]);
    }
    return 0;
}

```