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

def bfs():
    q = deque([1])
    while q:
        now = q.popleft()
        for nex,c,w in e[now]:
            if depth[nex]>= 0:
                continue
            depth[nex] = depth[now]+1
            dis[nex] = dis[now]+w
            lca[0][nex] = now
            q.append(nex)

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
        dfs1(nex,now,count1,count2)
        count1[c] -= 1
        count2[c] -= w
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))
size = n+1
bitlen = 19
lca = [[0]*size for i in range(bitlen)]
depth = [-1]*size
dis = [-1]*size
depth[1] = 0
dis[1] = 0

bfs()
for i in range(1,bitlen):
    for j in range(1,size):
        if lca[i-1][j] > 0:
            lca[i][j] = lca[i-1][lca[i-1][j]]
    
def search(x,y):
    if depth[x] > depth[y]:
        x,y = y,x
    for i in range(bitlen):
        if ((depth[y]-depth[x])>>i) & 1:
            y = lca[i][y]
    if x == y:
        return x
    for i in range(bitlen-1,-1,-1):
        if lca[i][x] != lca[i][y]:
            x = lca[i][x]
            y = lca[i][y]
    return lca[0][x]


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 = 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))

    
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)