Regular Meeting 2

A tree with $N$ vertices is given. The vertices are numbered from 1 to $N$. The $i$-th edge connects vertex $A_i$ and vertex $B_i$, and its weight is $C_i$. $(1 \leq i < N)$

The distance between two vertices is the maximum weight among the edges on the shortest path connecting them. The distance between a vertex and itself is 0.

People living in the tree hold $N$ meetings. The $i$-th meeting is attended by people living at vertices numbered from 1 to $i$. Because of the coronavirus situation this year, each meeting is held in person at $X$ places and continued online. Each meeting uses $X$ distinct vertices $v_1, \cdots, v_X$ of the tree. The vertices are chosen independently for each meeting. Once the vertices are chosen, each person travels to the one of $v_1, \cdots, v_X$ that requires the smallest travel distance.

The value of $X$ is one of the values from 1 to $K$. For each meeting, compute the sum, over $X = 1$ to $K$, of the minimum possible value of the maximum travel distance of the people, where the minimum is taken over the choice of vertices.

The first line contains the number of vertices $N$ and the integer $K$, separated by a space. $(1 \leq K \leq N \leq 300\,000)$

From the second line onward, the $N-1$ lines describe the edges. Line $i+1$ contains three integers $A_i$, $B_i$, and $C_i$, separated by spaces. This means there is an edge with weight $C_i$ connecting vertex $A_i$ and vertex $B_i$. $(1 \leq i < N)$ $(1 \leq A_i, B_i, C_i \leq N)$

Print $N$ lines. The $i$-th line contains the answer for the $i$-th meeting. $(1 \leq i \leq N)$