Problem: 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.

Example

Input

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


Output

130
200
60

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
from collections import defaultdict

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

depth = [0] * (N+1)
U = 17
ancestor = [[0] * (U+1) for _ in range(N+1)] # 2**i だけ遡った頂点の番号、行き過ぎは0

q = [(1,0,0)]
while q:
    qq = []
    for x,d,parent in q:
        depth[x] = d
        ax = ancestor[x]
        ax[0] = parent
        for i in range(d.bit_length() - 1):
            ax[i+1] = ancestor[ax[i]][i]
        for y, _, _ in graph[x]:
            if y == parent:
                continue
            qq.append((y, d+1, x))
    q = qq

def LCA(x,y):
    dx = depth[x]
    dy = depth[y]
    if dx > dy:
        x,y = y,x
        dx,dy = dy,dx
    diff = dy - dx
    while diff:
        step = diff & (-diff)
        y = ancestor[y][step.bit_length() - 1]
        diff -= step
    while x != y:
        j = 0
        while ancestor[x][j] != ancestor[y][j]:
            j += 1
        if j == 0:
            return ancestor[x][0]
        x = ancestor[x][j-1]
        y = ancestor[y][j-1]
    return x

tasks = [[] for _ in range(N+1)]
for i,(x,y,u,v) in enumerate(query):
    tasks[u].append((i,x,y,1))
    tasks[v].append((i,x,y,1))
    tasks[LCA(u,v)].append((i,x,y,-2))

answer = [0] * Q

def dfs(x = 1, sums = [0] * (N+1), nums = [0] * (N+1), total = 0, parent = 0):
    for i,c,d,coef in tasks[x]:
        answer[i] += coef * (total - sums[c] + nums[c] * d)
    for y,c,d in graph[x]:
        if y == parent:
            continue
        sums[c] += d
        nums[c] += 1
        total += d
        dfs(y, sums, nums, total, x)
        sums[c] -= d
        nums[c] -= 1
        total -= d
    return

dfs()

print('\n'.join(map(str,answer)))