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:
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
from pprint import pprint
from copy import deepcopy
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
from pprint import pprint


sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007


class LCA(object):
    def __init__(self, G, root=0):
        self.G = G
        self.root = root
        self.n = len(G)
        self.logn = (self.n - 1).bit_length()
        self.depth = [-1 if i != root else 0 for i in range(self.n)]
        self.parent = [[-1] * self.n for _ in range(self.logn)]
        self.dfs()
        self.doubling()

    def dfs(self):
        que = [self.root]
        while que:
            u = que.pop()
            for v in self.G[u]:
                if self.depth[v] == -1:
                    self.depth[v] = self.depth[u] + 1
                    self.parent[0][v] = u
                    que += [v]

    def doubling(self):
        for i in range(1, self.logn):
            for v in range(self.n):
                if self.parent[i - 1][v] != -1:
                    self.parent[i][v] = self.parent[i - 1][self.parent[i - 1][v]]

    def get(self, u, v):
        if self.depth[v] < self.depth[u]:
            u, v = v, u
        du = self.depth[u]
        dv = self.depth[v]

        for i in range(self.logn):  # depthの差分だけuを遡らせる
            if (dv - du) >> i & 1:
                v = self.parent[i][v]
        if u == v: return u  # 高さ揃えた時点で一致してたら終わり

        for i in range(self.logn - 1, -1, -1):  # そうでなければ上から二分探索
            pu, pv = self.parent[i][u], self.parent[i][v]
            if pu != pv:
                u, v = pu, pv
        return self.parent[0][u]
# query先読みで、頂点ごとに、どの色の長さを求めるのか。
n, q = LI()
edge_dict = [{} for _ in range(n)]
for u, v, c, d in LIR(n - 1):
    edge_dict[u - 1][v - 1] = (c, d)
    edge_dict[v - 1][u - 1] = (c, d)

lca = LCA(edge_dict)
query = [[] for _ in range(n)]
for q_idx, (x, y, u, v) in enumerate(LIR(q)):
    query[u - 1] += [(q_idx, x, y, 1)]
    query[v - 1] += [(q_idx, x, y, 1)]
    query[lca.get(u - 1, v - 1)] += [(q_idx, x, y, 0)]

def euler_tour(G, root=0):
    n = len(G)
    euler = []
    dq = deque([root])
    dq2 = deque()
    visited = [0] * n
    while dq:
        u = dq.pop()
        euler += [u]
        if visited[u]:
            continue
        for v in G[u]:
            if visited[v]:
                dq += [v]
            # [親頂点、子頂点、子頂点、。。。]と入れていく.その後連結
            else:
                dq2 += [v]
        dq.extend(dq2)
        dq2 = deque()
        visited[u] = 1
    return euler

dist = 0
color_len = [0] * n
color_cnt = [0] * n
ans = [0] * q
euler = euler_tour(edge_dict)
for i in range(1, len(euler)):
    now_v = euler[i]
    pre_v = euler[i - 1]
    c, d = edge_dict[pre_v][now_v]
    if lca.depth[pre_v] < lca.depth[now_v]:
        dist += d
        color_len[c] += d
        color_cnt[c] += 1
        for q_idx, x, y, flag in query[now_v]:
            if flag:
                ans[q_idx] += dist - color_len[x] + color_cnt[x] * y
            else:
                ans[q_idx] -= (dist - color_len[x] + color_cnt[x] * y) * 2
    else:
        dist -= d
        color_len[c] -= d
        color_cnt[c] -= 1


print(*ans, sep='\n')