Problem:
Score : 600 points

Problem Statement

There is a tree with N vertices numbered 1 to N.
The i-th edge in this tree connects Vertex a_i and Vertex b_i, and the color and length of that edge are c_i and d_i, respectively.
Here the color of each edge is represented by an integer between 1 and N-1 (inclusive). The same integer corresponds to the same color, and different integers correspond to different colors.

Answer the following Q queries:

Query j (1 \leq j \leq Q): assuming that the length of every edge whose color is x_j is changed to y_j, find the distance between Vertex u_j and Vertex v_j. (The changes of the lengths of edges do not affect the subsequent queries.)

Constraints

2 \leq N \leq 10^5

1 \leq Q \leq 10^5

1 \leq a_i, b_i \leq N

1 \leq c_i \leq N-1

1 \leq d_i \leq 10^4

1 \leq x_j \leq N-1

1 \leq y_j \leq 10^4

1 \leq u_j < v_j \leq N

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N Q
a_1 b_1 c_1 d_1
:
a_{N-1} b_{N-1} c_{N-1} d_{N-1}
x_1 y_1 u_1 v_1
:
x_Q y_Q u_Q v_Q

Output

Print Q lines. The j-th line (1 \leq j \leq Q) should contain the answer to Query j.

Sample Input 1

5 3
1 2 1 10
1 3 2 20
2 4 4 30
5 2 1 40
1 100 1 4
1 100 1 5
3 1000 3 4

Sample Output 1

130
200
60

The graph in this input is as follows:

Here the edges of Color 1 are shown as solid red lines, the edge of Color 2 is shown as a bold green line, and the edge of Color 4 is shown as a blue dashed line.

Query 1: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 4 is 100 + 30 = 130.

Query 2: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 5 is 100 + 100 = 200.

Query 3: Assuming that the length of every edge whose color is 3 is changed to 1000 (there is no such edge), the distance between Vertex 3 and Vertex 4 is 20 + 10 + 30 = 60. Note that the edges of Color 1 now have their original lengths.

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import bisect


def euler_tour(tree: list, root: int):
    """根をrootとしたときの辺に対するオイラーツアーを行う
    posの部分木が区間[begin[pos], end[pos])に対応する
    """
    n = len(tree)
    trail = []
    weight = []
    color = []
    begin = [-1] * n
    end = [-1] * n
    par = {root: -1}
    stack = [(root, 0, 0)]

    while stack:
        v, c, col = stack.pop()
        if v >= 0:  # 行きがけ順の処理
            trail.append(v)
            weight.append(c)
            color.append(col)
            begin[v] = len(trail) - 1
            end[v] = len(trail)
            stack.append((~v, -c, col))
            for nxt_v, nxt_c, nxt_col in tree[v]:
                if nxt_v not in par:
                    par[nxt_v] = v
                    stack.append((nxt_v, nxt_c, nxt_col))

        else:  # 帰りがけ順の処理
            if ~v == root:
                continue
            trail.append(par[~v])
            weight.append(c)
            color.append(col)
            end[par[~v]] = len(trail)

    return trail, weight, color, begin, end


n, q = map(int, input().split())
info = [list(map(int, input().split())) for i in range(n - 1)]
query = [list(map(int, input().split())) for i in range(q)]

tree = [[] for i in range(n)]
for a, b, col, dist in info:
    a -= 1
    b -= 1
    tree[a].append((b, dist, col))
    tree[b].append((a, dist, col))

trail, weight, color, begin, end = euler_tour(tree, 0)

# 辺の重みを色ごとに累積和する
w_col = [[0] for i in range(n)]
for i in range(len(weight)):
    col = color[i]
    w_col[col].append(weight[i] + w_col[col][-1])

# 辺の個数を色ごとに累積和する
cnt_col = [[0] for i in range(n)]
for i in range(len(weight)):
    col = color[i]
    cnt_col[col].append(int(2 * (weight[i] >= 1)) - 1 + cnt_col[col][-1])

# 色ごとのインデックス
ind = [[] for i in range(n)]
for i in range(len(weight)):
    col = color[i]
    ind[col].append(i)

# 辺の重みを累積和にする
for i in range(len(weight) - 1):
    weight[i + 1] += weight[i]

for x, y, u, v in query:
    u -= 1
    v -= 1
    ans = weight[begin[u]] + weight[begin[v]]
    l = bisect.bisect_right(ind[x], begin[u])
    r = bisect.bisect_right(ind[x], begin[v])
    print(ans)