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
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

n,q = map(int,readline().split())
data = list(map(int,read().split()))
abcd = data[:4*(n-1)]
xyuv = data[4*(n-1):]

root = 1
dbl_max = 20
dbl = [[0] * dbl_max for _ in range(n+1)]
depth = [0] * (n+1)
links = [[] for _ in range(n+1)]
it = iter(abcd)
for a,b,c,d in zip(it,it,it,it):
    links[a].append((b,c,d))
    links[b].append((a,c,d))

stack = [root]
while(stack):
    i = stack.pop()
    for j,c,d in links[i]:
        if(j==dbl[i][0]):
            continue
        dbl[j][0] = i
        depth[j] = depth[i]+1
        stack.append(j)

for i in range(1,dbl_max):
    for j in range(1,n+1):
        dbl[j][i] = dbl[dbl[j][i-1]][i-1]

def get_lca(x,y):
    if(depth[x] > depth[y]):
        x,y = y,x

    dif = depth[y]-depth[x]
    for i in range(dbl_max):
        if(dif==0):
            break
        if(dif>>i)&1:
            y = dbl[y][i]
            dif -= (1<<i)

    for i in range(dbl_max-1,-1,-1):
        if(dbl[x][i] != dbl[y][i]):
            x = dbl[x][i]
            y = dbl[y][i]

    if(x!=y):
        return dbl[x][0]
    else:
        return x

ans = [0] * q
query = [[] for _ in range(n+1)]
for i in range(q):
    x,y,u,v = xyuv[i*4:i*4+4]

    lca = get_lca(u,v)
    query[u].append((i,x,y,1))
    query[v].append((i,x,y,1))
    query[lca].append((i,x,y,-2))

tot = 0
c_cost = [[0,0] for _ in range(n)]
stack = []
for j,c,d in links[root]:
    stack.append((root,j,c,d,1))

while(stack):
    a,b,c,d,num = stack.pop()
    tot += d*num
    c_cost[c][0] += d*num
    c_cost[c][1] += num

    if(num==1):
        for qi,x,y,num_q in query[b]:
            ans[qi] += num_q * (tot - c_cost[x][0] + c_cost[x][1]*y)

        next = []
        for j,c,d in links[b]:
            if(j==dbl[b][0]):
                stack.append((b,j,c,d,-1))
            else:
                next.append((b,j,c,d,1))
        stack += next
    # print(c_cost)

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