#include <iostream>
#include <cstring>
using namespace std;

const int N = 110, M = 2 * N;
int n, m;
int h[N], e[M], ne[M], w[M], idx;
int f[N][N];

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

void dfs(int u, int parent) {
    for (int i = h[u]; i != -1; i = ne[i]) {
        int v = e[i];
        if (v == parent) continue;
        
        // 先递归求解f[v]
        dfs(v, u);

		// 枚举树枝总数的上界（即保留不超过多少个树枝）
        for (int j = m; j >= 1; j--) 
        	// 枚举v这棵子树取多少个树枝
            for (int k = 0; k < j; k++)
                f[u][j] = max(f[u][j], f[u][j - k - 1] + f[v][k] + w[i]);
    }
}

int main() {
    cin >> n >> m;

    memset(h, -1, sizeof h);
    for (int i = 0; i < n - 1; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        // 只知道树根但不知道边的方向，所以只能建双向边
        add(a, b, c), add(b, a, c);
    }

    dfs(1, -1);

    cout << f[1][m] << endl;

    return 0;
}