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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

# verified: ABC014-D
# calculate LCA O(logN) for each query

# kprev[k][i]: 2^k parent of node i 
# if there is no node returns "None"
# Depth[i]: Depth of node i from rootNode

class Doubling():
    def __init__(self, graph, Depth, prev, rootNode=0):
        self.graph = graph
        self.N = len(graph)
        self.Log_N = (self.N-1).bit_length()
        self.rootNode = rootNode
        # BFSしない
        self.prev = prev 
        self.Depth = Depth

        # construct
        self.kprev = [self.prev]
        S = self.prev
        for k in range(self.Log_N):
            T = [None]*self.N
            for i in range(self.N):
                if S[i] is None:
                    continue
                T[i] = S[S[i]]
            self.kprev.append(T)
            S = T

    # LCA
    def LCAquery(self, u, v):
        dd = self.Depth[v] - self.Depth[u]
        if dd < 0:
            u, v = v, u
            dd = -dd
        
        # set same Depth
        for k in range(self.Log_N+1):
            if dd & 1:
                v = self.kprev[k][v]
            dd >>= 1
        
        if u == v:
            return u
        
        for k in reversed(range(self.Log_N)):
            pu = self.kprev[k][u]
            pv = self.kprev[k][v]
            if pu != pv:
                u, v = pu, pv
        
        return self.kprev[0][u]

import sys
input = sys.stdin.readline
# sys.setrecursionlimit(10**6)
from bisect import bisect_right

N, Q = map(int, input().split())
graph = [[] for _ in range(N)]
for _ in range(N-1):
    a, b, c, d = map(int, input().split())
    graph[a-1].append((b-1, c, d))
    graph[b-1].append((a-1, c, d))

Query = [list(map(int, input().split())) for _ in range(Q)]


D = [-1]*N
D[0] = 0
EulerTour = [(0, +1)]
Color = [0]*N
Weight = [0]*N
PointFront = [0]*N
prev = [None]*N
Depth = [0]*N
# def dfs(p):
#     PointFront[p] = len(EulerTour)
#     EulerTour.append((p, True))
#     for np, c, d in graph[p]:
#         if D[np] == -1:
#             Color[np] = c
#             Weight[np] = d
#             D[np] = D[p] + d
#             dfs(np)
#     EulerTour.append((p, False))

def dfs2():
    stack = [0]
    while stack:
        p = stack[-1]
        for np, c, d in graph[p]:
            if D[np] == -1:
                PointFront[np] = len(EulerTour)
                EulerTour.append((np, +1))
                Color[np] = c
                Weight[np] = d
                D[np] = D[p] + d
                Depth[np] = Depth[p] + 1
                prev[np] = p
                stack.append(np)
                break
        if p == stack[-1]:
            EulerTour.append((p, -1))    
            stack.pop()
            

dfs2()


Indexes = [[] for _ in range(N)]
Weights = [[0] for _ in range(N)]
Colors = [[0] for _ in range(N)]
for i, (p, num) in enumerate(EulerTour):
    if p == 0: continue
    w = Weight[p]*num
    c = Color[p]
    Weights[c].append(Weights[c][-1]+w)
    Colors[c].append(Colors[c][-1]+num)
    Indexes[c].append(i)



# point までにある color 指定された時のスコア
def fixed_score(point, color, replace):
    ind = bisect_right(Indexes[color], PointFront[point])
    return D[point] - Weights[color][ind] + Colors[color][ind]*replace


def main():
    doubling = Doubling(graph, Depth, prev)
    ans = []
    for x, y, u, v in Query:
        u -= 1; v -= 1
        lca = doubling.LCAquery(u, v)
        ans.append(fixed_score(u, x, y) + fixed_score(v, x, y) - 2*fixed_score(lca, x, y))
    print(*ans, sep="\n")

if __name__ == "__main__":
    main()