#include <bits/stdc++.h>
using namespace std;

#define dump(x) cout << (x) << '\n'
#define Int int64_t
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()

Int INF = 1e18;
int inf = 1e9;
Int mod = 1e9+7;

Int par[100001];
Int sz[100001];

Int root(Int x) {
    if (x == par[x]) return x;
    return par[x] = root(par[x]);
}

void unite(Int x, Int y) {
    Int rx = root(x);
    Int ry = root(y);
    if (rx == ry) return;
    par[ry] = rx;
    sz[rx] += sz[ry];
}

int main() {
    vector<Int> res;
    Int n, m;
    while (cin >> n >> m and n) {
        vector<pair<Int, pair<Int, Int>>> G(m);
        for (Int i = 0; i < m; i++) {
            cin >> G[i].se.fi >> G[i].se.se >> G[i].fi;
        }
        for (Int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
        sort(all(G));
        vector<Int> mst;
        for (Int i = 0; i < m; i++) {
            pair<Int, Int> p = G[i].se;
            if (root(p.fi) != root(p.se)) {
                mst.push_back(G[i].fi);
                unite(p.fi, p.se);
            }
        }
        sort(all(mst));
        res.push_back(mst[(n - 1) / 2]);
    }
    for (Int i = 0; i < (Int)res.size(); i++) {
        dump(res[i]);
    }
    return 0;
}


### Prompt

Please create a solution in Cpp 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;

#define dump(x) cout << (x) << '\n'
#define Int int64_t
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()

Int INF = 1e18;
int inf = 1e9;
Int mod = 1e9+7;

Int par[100001];
Int sz[100001];

Int root(Int x) {
    if (x == par[x]) return x;
    return par[x] = root(par[x]);
}

void unite(Int x, Int y) {
    Int rx = root(x);
    Int ry = root(y);
    if (rx == ry) return;
    par[ry] = rx;
    sz[rx] += sz[ry];
}

int main() {
    vector<Int> res;
    Int n, m;
    while (cin >> n >> m and n) {
        vector<pair<Int, pair<Int, Int>>> G(m);
        for (Int i = 0; i < m; i++) {
            cin >> G[i].se.fi >> G[i].se.se >> G[i].fi;
        }
        for (Int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
        sort(all(G));
        vector<Int> mst;
        for (Int i = 0; i < m; i++) {
            pair<Int, Int> p = G[i].se;
            if (root(p.fi) != root(p.se)) {
                mst.push_back(G[i].fi);
                unite(p.fi, p.se);
            }
        }
        sort(all(mst));
        res.push_back(mst[(n - 1) / 2]);
    }
    for (Int i = 0; i < (Int)res.size(); i++) {
        dump(res[i]);
    }
    return 0;
}


```