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**7)

N, Q = map(int, input().split())
edge = [[] for _ in range(N+1)]
for _ in range(N-1):
        a, b, c, d = map(int, input().split())
        edge[a].append((b, c, d))
        edge[b].append((a, c, d))
        
depth = [0] * (N+1)
U = 17
ancestor = [[0] * (U+1) for _ in range(N+1)]

q = [(1, 0, 0)]
while q:
        qq = []
        for x, d, parent in q:
                depth[x] = d
                ax = ancestor[x]
                ax[0] = parent
                for i in range(d.bit_length()-1):
                        ax[i+1] = ancestor[ax[i]][i]

                for y, _, _ in edge[x]:
                        if y == parent:
                                continue
                        qq.append((y, d+1, x))
        q = qq

def lca(x, y):
        dx = depth[x]
        dy = depth[y]
        if dx > dy:
                x, y = y, x
                dx, dy = dy, dx
        diff = dy - dx
        while diff:
                step = diff & (-diff)
                y = ancestor[y][step.bit_length()-1]
                diff -= step
        while x!=y:
                j = 0
                while ancestor[x][j] != ancestor[y][j]:
                        j += 1
                if j==0:
                        return ancestor[x][0]
                x = ancestor[x][j-1]
                y = ancestor[y][j-1]
        return x

task = [[] for _ in range(N+1)]
for i in range(Q):
        x, y, u, v = map(int, input().split())
        task[u].append((i, x, y, 1))
        task[v].append((i, x, y, 1))
        task[lca(u, v)].append((i, x, y, -2))

answer = [0] * Q

def dfs(x=1, sums=[0]*(N+1), nums=[0]*(N+1), total=0, parent=0):
        for i, c, d, coef in task[x]:
                answer[i] += coef * (total - sums[c] + nums[c]*d)
        for y, c, d in edge[x]:
                if y == parent:
                        continue
                sums[c] += d
                nums[c] += 1
                total += d
                dfs(y, sums, nums, total, x)
                sums[c] -= d
                nums[c] -= 1
                total -= d
        return

dfs()

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