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: Time Limit Exceeded)

import sys
sys.setrecursionlimit(100000000)
def input():
    return sys.stdin.readline()[:-1]
from math import log

class LCA:
    def __init__(self, N, root, es):
        self.N = N
        self.root = root
        self.es = es
        self.LOG_N = LOG_N = int(log(N, 2)) + 1
        self.parent = [[-1 for _ in range(N)] for _ in range(LOG_N)]
        self.depth = [0 for _ in range(N)]
        self.dfs(root, -1, 0)
        parent = self.parent
        for k in range(LOG_N-1):
            for v in range(N):
                if parent[k][v] < 0:
                    parent[k+1][v] = -1
                else:
                    parent[k+1][v] = parent[k][parent[k][v]]
        self.parent = parent

    def dfs(self, v, p, d):
        self.parent[0][v] = p
        self.depth[v] = d
        for u in self.es[v]:
            if u[0] != p:
                self.dfs(u[0], v, d+1)

    def lca(self, u, v):
        LOG_N = self.LOG_N
        depth = self.depth
        parent = self.parent
        if depth[u] > depth[v]:
            t, u = u, v
            v = t
        for k in range(LOG_N):
            if (depth[v] - depth[u]) >> k & 1:
                v = parent[k][v]
        if u == v:
            return u
        for k in range(LOG_N)[::-1]:
            if parent[k][u] != parent[k][v]:
                u, v = parent[k][u], parent[k][v]
        return parent[0][u]

N, Q = map(int, input().split())
es = [[] for i in range(N+1)]
for i in range(N-1):
    a, b, c, d = map(int, input().split())
    es[a].append([b, c, d])
    es[b].append([a, c, d])
lca = LCA(N+1, 1, es)
qs = [[] for i in range(N+1)]
query = [None] * Q
for i in range(Q):
    x, y, u, v = map(int, input().split())
    qs[u].append([i, x, 1])
    qs[v].append([i, x, 1])
    c = lca.lca(u, v)
    qs[c].append([i, x, -2])
    query[i] = [y, 0, 0]
ds = [0] * N
cnt = [0] * N
def EulerTour(v, p = -1, alld=0):
    for q in qs[v]:
        id, col, f = q
        query[id][1] += f * cnt[col]
        query[id][2] += f * (alld - ds[col])
    for u, c, d in es[v]:
        if u != p:
            ds[c] += d
            cnt[c] += 1
            EulerTour(u, v, alld+d)
            ds[c] -= d
            cnt[c] -= 1

EulerTour(1)
for y, c, d in query:
    print(y * c + d)