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
from math import log2
import sys
sys.setrecursionlimit(10**6)
input=sys.stdin.readline

def f(n,q):
    def dfs(i, oi=-1, dpt=0):
        depth[i] = dpt
        parent_2k[0][i] = oi
        for ki, c, d in to[i]:
            if ki == oi: continue
            dfs(ki, i, dpt + 1)
        return

    def make_parent(level):
        parent_2kk1 = parent_2k[0]
        for k in range(1, level):
            parent_2kk = parent_2k[k]
            for i in range(n):
                parent1 = parent_2kk1[i]
                if parent1 != -1:
                    parent_2kk[i] = parent_2kk1[parent1]
            parent_2kk1 = parent_2kk
        return

    def lca(u, v):
        dist_uv = depth[u] - depth[v]
        if dist_uv < 0:
            u, v = v, u
            dist_uv *= -1

        k = 0
        while dist_uv:
            if dist_uv & 1:
                u = parent_2k[k][u]
            dist_uv >>= 1
            k += 1

        if u == v:
            return u

        for k in range(int(log2(depth[u])) + 1, -1, -1):
            pu = parent_2k[k][u]
            pv = parent_2k[k][v]
            if pu != pv:
                u = pu
                v = pv
        return parent_2k[0][u]

    to = defaultdict(list)
    root = 0
    depth = [0] * n
    level = int(log2(n)) + 1
    parent_2k = [[-1] * n for _ in range(level)]

    for _ in range(n - 1):
        a, b, c, d = list(map(int, input().split()))
        to[a - 1] += [[b - 1, c - 1, d]]
        to[b - 1] += [[a - 1, c - 1, d]]

    dfs(root)
    make_parent(level)

    qs = [[] for _ in range(n)]
    for qi in range(q):
        x, y, u, v = map(int, input().split())
        qs[u - 1].append([qi, x - 1, y, 1])
        qs[v - 1].append([qi, x - 1, y, 1])
        qs[lca(u - 1, v - 1)].append([qi, x - 1, y, -2])

    cnt_c = [0] * n
    sum_c = [0] * n
    ans = [0] * q

    def dfs2(i, oi=-1, sum_d=0):
        for qi, qc, qd, qk in qs[i]:
            ans[qi] += (sum_d - sum_c[qc] + qd * cnt_c[qc]) * qk
        for ki, c, d in to[i]:
            if ki == oi: continue
            cnt_c[c] += 1
            sum_c[c] += d
            dfs2(ki, i, sum_d + d)
            cnt_c[c] -= 1
            sum_c[c] -= d
        return

    dfs2(root)
    for x in ans:
        print(x)

n, q = map(int, input().split())
f(n,q)