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)

import sys
from collections import defaultdict
sys.setrecursionlimit(10**7)
readline = sys.stdin.readline
 
N, Q = map(int, readline().split())
G = [[] for _ in range(N)]
dp = {}

def init():
    for v in range(N-1):
        a, b, c, d = map(int, readline().split())
        a, b = a-1, b-1
        G[a].append(b)
        G[b].append(a)
        dp[(min(a,b), max(a,b))] = c, d

init()
S = [0]*(2*N)
F = [0]*N
depth = [0]*N
depthh = [0]*N
mmn = [[0] for _ in range(N)]
msn = [[0] for _ in range(N)]
iin = [[] for _ in range(N)]
w2i = [0]*(2*N)
ii = -1
def dfs(bb, v, d, dd, cc, dpd):
    global ii
    ii += 1
    F[v] = ii
    depth[v] = d
    depthh[v] = dd
    S[ii] = v
    mmn[cc].append(mmn[cc][-1]+1)
    msn[cc].append(msn[cc][-1]+dpd)
    iin[cc].append(v)
    w2i[v] = ii
    for w in G[v]:
        if w == bb:
            continue
        cc, dpd = dp[(min(v,w), max(v,w))]
        dfs(v, w, d+1, dd+dpd, cc, dpd)
#        bb, v, d, dd = v, w, d+1, dd+dpd
        ii += 1
        S[ii] = v
        mmn[cc].append(mmn[cc][-1]-1)
        msn[cc].append(msn[cc][-1]-dpd)
        iin[cc].append(-w)
        w2i[-w] = ii


dfs(None, 0, 0, 0, 0,0)
lnc = [len(x)-1 for x in mmn]
 
INF = (N, None)
M = 2*N
M0 = 2**(M-1).bit_length()
data = [INF]*(2*M0)

def mk_data():
    for i, v in enumerate(S):
        data[M0-1+i] = (depth[v], i)
    for i in range(M0-2, -1, -1):
        data[i] = min(data[2*i+1], data[2*i+2])

mk_data()
 
def _query(a, b):
    yield INF
    a += M0; b += M0
    while a < b:
        if b & 1:
            b -= 1
            yield data[b-1]
        if a & 1:
            yield data[a-1]
            a += 1
        a >>= 1; b >>= 1
 
def query(u, v):
    fu = F[u]; fv = F[v]
    if fu > fv:
        fu, fv = fv, fu
    return S[min(_query(fu, fv+1))[1]]

def b_s(ll, mm, ms, ind, lw, hw, target_i, lncc):
    if target_i < w2i[ll[0]]:
        return 0, 0
    if target_i >= w2i[ll[lncc]]:
        return 0, 0
    for _ in range(N):
        if w2i[ll[ind]] <= target_i:
            if target_i < w2i[ll[ind+1]]:
                return (mm[ind+1], ms[ind+1])
            ind, lw = (ind+1 + hw)//2, ind+1
            continue
        ind, hw = (ind-1 + lw)//2, ind-1

def iro(u, v, cc, x, y):
    u_n, u_d = b_s(iin[x], mmn[x], msn[x], (lnc[x]-1)//2, 0, lnc[x]-1, w2i[u], lnc[x]-1)
    v_n, v_d = b_s(iin[x], mmn[x], msn[x], (lnc[x]-1)//2, 0, lnc[x]-1, w2i[v], lnc[x]-1)
    c_n, c_d = b_s(iin[x], mmn[x], msn[x], (lnc[x]-1)//2, 0, lnc[x]-1, w2i[cc], lnc[x]-1)
    return (u_n + v_n - 2 * (c_n)) * y - (u_d + v_d - 2 * (c_d))

def process():
    for q in range(Q):
        x, y, u, v = map(int, readline().split())
        u, v = u-1, v-1
        cc = query(u, v)
        if lnc[x] == 0:
            diff = 0
        else:
            diff = iro(u, v, cc, x, y)
        print(depthh[u] + depthh[v] - 2*depthh[cc] + diff)

process()