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)

from collections import defaultdict
import math

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

depth = [0] * (N+1)
dist_sum = [0] * (N+1)
parent = list(range(N+1))

#dfs
stack = []
stack.append((1, 1, 0, 0))
while stack:
        now, prev, dep, dist = stack.pop()
        depth[now] = dep
        parent[now] = prev
        dist_sum[now] = dist
        for e, c, d in edge[now]:
                if e == prev:
                        continue
                else:
                        stack.append((e, now, dep+1, dist+d))

U = 17
doubling = [[1] * U for i in range(N+1)]
for i in range(U):
        for j in range(1, N+1):
                doubling[j][i] = parent[j]
        parent = [parent[parent[i]] for i in range(N+1)]

def lca(a, b):
        x, y = depth[a], depth[b]
        if x > y:
                x, y = y, x
                a, b = b, a        
        diff = y-x
        if diff > 0:
                for i in reversed(range(U)):
                        if diff > 2**i:
                                b = doubling[b][i]
                                diff -= 2**i
                b = doubling[b][0]
        if a == b:
                return a
        
        for i in reversed(range(int(math.log2(x)) + 2)):
                if doubling[a][i] != doubling[b][i]:
                        a, b = doubling[a][i], doubling[b][i]
        return doubling[a][0]

task = defaultdict(list)
for i in range(Q):
        x, y, u, v = map(int, input().split())
        task[u].append((i, x, y, 1))
        task[v].append((i, x, y, 1))
        task[lca(u, v)].append((i, x, y, -2))

stack = []
ans = [0] * Q
color_num = [0] * N
color_dist = [0] * N
stack.append((1, -1, color_num, color_dist))
while stack:
        now, prev, color_num, color_dist = stack.pop()
        if len(task[now]) > 0:
                for i, x, y, coeff in task[now]:
                        ans[i] += coeff * (dist_sum[now] - color_dist[x] + color_num[x] * y)
                        
        for e, c, d in edge[now]:
                if e == prev:
                        continue
                color_num[c] += 1
                color_dist[c] += d
                stack.append((e, now, color_num[:], color_dist[:]))
                color_num[c] -= 1
                color_dist[c] -= d

print('\n'.join([str(ans[i]) for i in range(Q)]))