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
sys.setrecursionlimit(10**5+5)
from collections import defaultdict,deque
input = sys.stdin.readline
class LCA:

    def __init__(self,n):
        self.size = n+1
        self.bitlen = n.bit_length()
        self.lca = [[0]*self.size for i in range(self.bitlen)]
        self.depth = [-1]*self.size
        self.dis = [-1]*self.size
        self.depth[1] = 0
        self.dis[1] = 0
    def make(self,root):
        q = deque([root])
        while q:
            now = q.popleft()
            for nex,c,w in e[now]:
                if self.depth[nex]>= 0:
                    continue
                self.depth[nex] = self.depth[now]+1
                self.dis[nex] = self.dis[now]+w
                self.lca[0][nex] = now
                q.append(nex)
        for i in range(1,self.bitlen):
            for j in range(self.size):
                if self.lca[i-1][j] > 0:
                    self.lca[i][j] = self.lca[i-1][self.lca[i-1][j]]
    
    def search(self,x,y):
        dx = self.depth[x]
        dy = self.depth[y]
        if dx < dy:
            x,y = y,x
            dx,dy = dy,dx
        dif = dx-dy
        while dif:
            s = dif & (-dif)
            x = self.lca[s.bit_length()-1][x]
            dif -= s
        if x == y:
            return x
        for i in range(self.bitlen-1,-1,-1):
            if self.lca[i][x] != self.lca[i][y]:
                x = self.lca[i][x]
                y = self.lca[i][y]
        
        return self.lca[0][x]

n,q = map(int,input().split())
e = [[] for i in range(n+1)]
for i in range(n-1):
    a,b,c,d = map(int,input().split())
    e[a].append((b,c,d))
    e[b].append((a,c,d))
    
lca = LCA(n)
lca.make(1)

Q = []
dcount = defaultdict(lambda : defaultdict(int))
dweight = defaultdict(lambda : defaultdict(int))
for i in range(q):
    x,y,u,v = map(int,input().split())
    a = lca.search(u,v)
    dcount[u][x] = 0
    dcount[v][x] = 0
    dcount[a][x] = 0
    dweight[u][x] = 0
    dweight[v][x] = 0
    dweight[a][x] = 0
    Q.append((x,y,u,v,a))

dis = [0]*(n+1)


def dfs1(now,p,count1,count2):
    for cou in dcount[now]:
        dcount[now][cou] = count1[cou]
    for wei in dweight[now]:
        dweight[now][wei] = count2[wei]

    for nex,c,w in e[now]:
        if nex == p:
            continue
        count1[c] += 1
        count2[c] += w
        dis[nex] =dis[now]+w
        dfs1(nex,now,count1,count2)
        count1[c] -= 1
        count2[c] -= w
    
dfs1(1,0,defaultdict(int),defaultdict(int))

for x,y,u,v,a in Q:
    cal = dis[u]+dis[v]-2*dis[a]
    cal += y*(dcount[u][x]+dcount[v][x]-2*dcount[a][x]) - dweight[u][x]-dweight[v][x]+2*dweight[a][x]
    print(cal)