Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by natural numbers the root of binary apple tree, points of branching and the ends of twigs. This way we may distinguish different branches by their ending points. We will assume that root of tree always is numbered by 1 and all numbers used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:

   2   5
\ /
3   4
\ /
1

Input

First line of input contains two numbers: N and Q (1 ≤ Q ≤ N, 1 < N ≤ 100). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. Next N-1 lines contains descriptions of branches. Each description consists of a three integer numbers divided by spaces. The first two of them define branch by it's ending points. The third number defines the number of apples on this branch. You may assume that no branch contains more than 30000 apples.

Output

Output file should contain the only number — amount of apples that can be preserved. And don't forget to preserve tree's root ;-)

Sample

5 2
            1 3 1
            1 4 10
            2 3 20
            3 5 20
            

21
            

#include <iostream>
using namespace std;

const int MAXN = 110;
const int MAXQ = 110;

int n, q;
int d[MAXN][MAXQ];
int g[MAXN][MAXN], deg[MAXN], nw[MAXN], son[MAXN][2];
int root;
int f[MAXN], isleaf[MAXN];

inline int maxt(int a, int b) {
    return a > b ? a : b;
}

void DFS(int v) {
    int i, t = 0;
    for (i=1; i<=n; i++) {
        if (g[v][i] != -1 && !f[i]) {
            son[v][t++] = i;
            f[i] = 1;
            nw[i] = g[v][i];
            DFS(i);
        }
    }
    if (!t) isleaf[v] = 1;
}

void buildT() {
    int i;
    for (i=1; i<=n; i++) {
        if (deg[i] == 2) {
            root = i;
            break;
        }
    }
    f[root] = 1; nw[root] = 0;
    DFS(root);
}

int DP(int ti, int tl) {
    if (tl <= 0) return 0;
    if (d[ti][tl] != -1) return d[ti][tl];
    if (isleaf[ti]) return nw[ti];
    int k, tmp = 0;
    for (k=0; k<tl; k++) {
        tmp = maxt(tmp, DP(son[ti][0], k)+DP(son[ti][1], tl-1-k));
    }
    d[ti][tl] = tmp + nw[ti];
    return d[ti][tl];
}

int main() {
    int i, j, k, x, y, w;
    scanf("%d%d", &n, &q);
    memset(g, -1, sizeof(g));
    memset(d, -1, sizeof(d));
    for (i=0; i<n-1; i++) {
        scanf("%d%d%d", &x, &y, &w); 
        g[x][y] = g[y][x] = w;
        deg[x]++, deg[y]++;
    }
    buildT();
    printf("%d\n", DP(root, q+1));

    return 0;
}