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

input=sys.stdin.readline
sys.setrecursionlimit(10**7)

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

prv=[-1]*N
depth=[0]*N
dist=[0 for i in range(N)]
def dfs(v,pv):
    prv[v]=pv
    for nv,color,weight in edge[v]:
        if nv!=pv:
            depth[nv]=depth[v]+1
            dist[nv]=dist[v]+weight
            dfs(nv,v)

dfs(0,-1)

LV = (N-1).bit_length()
def construct(prv):
    kprv = [prv]
    S = prv
    for k in range(LV):
        T = [0]*N
        for i in range(N):
            if S[i] is None:
                continue
            T[i] = S[S[i]]
        kprv.append(T)
        S = T
    return kprv

kprv=construct(prv)

def lca(u, v):
    dd = depth[v] - depth[u]
    if dd < 0:
        u, v = v, u
        dd = -dd

    # assert depth[u] <= depth[v]
    for k in range(LV+1):
        if dd & 1:
            v = kprv[k][v]
        dd >>= 1

    # assert depth[u] == depth[v]
    if u == v:
        return u

    for k in range(LV-1, -1, -1):
        pu = kprv[k][u]; pv = kprv[k][v]
        if pu != pv:
            u = pu; v = pv

    # assert kprv[0][u] == kprv[0][v]
    return kprv[0][u]

query=[]
q=[[] for i in range(N)]
data={}
for i in range(Q):
    x,y,u,v=map(int,input().split())
    LCA=lca(u-1,v-1)
    query.append((x,y,u-1,v-1,LCA))
    q[u-1].append(x)
    q[v-1].append(x)
    q[LCA].append(x)

num=[0 for i in range(N)]
cdist=[0 for i in range(N)]

def dfsc(v,pv,color,weight):
    pn=num[color]
    num[color]+=1
    pd=cdist[color]
    cdist[color]+=weight
    for x in q[v]:
        data[(v,x)]=(0,0)
        data[(v,x)]=(num[x],cdist[x])

    for nv,ncolor,nweight in edge[v]:
        if nv!=pv:
            dfsc(nv,v,ncolor,nweight)

    num[color]=pn
    cdist[color]=pd



dfsc(0,0,0,0)

for x,y,u,v,LCA in query:
    d1=dist[LCA]-data[(LCA,x)][1]+data[(LCA,x)][0]*y
    d2=dist[u]-data[(u,x)][1]+data[(u,x)][0]*y
    d3=dist[v]-data[(v,x)][1]+data[(v,x)][0]*y
    print(d2-d1+d3-d1)