Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

130
200
60

The graph in this input is as follows:

Here the edges of Color 1 are shown as solid red lines, the edge of Color 2 is shown as a bold green line, and the edge of Color 4 is shown as a blue dashed line.

Query 1: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 4 is 100 + 30 = 130.

Query 2: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 5 is 100 + 100 = 200.

Query 3: Assuming that the length of every edge whose color is 3 is changed to 1000 (there is no such edge), the distance between Vertex 3 and Vertex 4 is 20 + 10 + 30 = 60. Note that the edges of Color 1 now have their original lengths.

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N, Q = gets.chomp.split(" ").map(&:to_i)
E = Hash.new{|h, k| h[k] = []}
C = Hash.new{|h, k| h[k] = []}
(N - 1).times{|n|
  a, b, c, d = gets.chomp.split(" ").map(&:to_i)
  E[a] << [b, c, d, n]
  E[b] << [a, c, d, n]
}
M = []
Q.times{
  M << gets.chomp.split(" ").map(&:to_i)
}

D = Hash.new{|h, k| h[k] = Array.new}
stk = []
V = Hash.new{|h, k| h[k] = [0, 0, []]}
stk.push([1, 0, 0, 0, 0])
V[1] = [0, 0, 0, []]
until(stk.empty?) do
  xv, xc, xdp, xd, dd = stk.pop
  D[xdp] << xv
  C[xc] << [xv, dd]
  E[xv].each{|a, c, d|
    next if V.include?(a)
    V[a] = [xd + d, c, xdp + 1, []]
    dpth = 1
    while(xdp + 1 - dpth >= 0) do
      V[a][3] << D[xdp + 1 - dpth][-1]
      dpth <<= 1
    end
    stk.push([a, c, xdp + 1, xd + d, d])
  }
end

def get_upper_depth(a, dpth)
  df = V[a][2] - dpth
  return a if df == 0
  i = 0
  while(2 ** i < df) do
    i += 1
  end
  return V[a][3][i] if 2 ** i == df
  return get_upper_depth(V[a][3][i-1], dpth)
end

def get_same_depth(a, b)
  da = V[a][2]
  db = V[b][2]
  return get_same_depth(a, get_upper_depth(b, da)) if da < db
  return get_same_depth(b, get_upper_depth(a, db)) if da > db
  return [V[a][0], V[a][2]] if V[a][0] == V[b][0]
  return [V[a][3][0], V[a][2] - 1] if V[a][3][0] == V[b][3][0]
  return [V[a][3][1], V[a][2] - 2] if V[a][3][1] == V[b][3][1]
  (2...V[a][3].size).each{|i|
    return get_same_depth(V[a][3][i - 1], V[b][3][i - 1]) if V[a][3][i] == V[b][3][i]
  }
  return get_same_depth(V[a][3][-1], V[b][3][-1])
end

M.each{|x, y, u, v|
  ans = V[u][0] + V[v][0]
  ma, md = get_same_depth(u, v)
  ans -= (V[ma][0] * 2)
  C[x].each{|c, dd|
    dc = V[c][2]
    if md < dc && dc <= V[u][2] then
      ans += (y - dd) if c == get_upper_depth(u, dc)
    end
    if md < dc && dc <= V[v][2] then
      ans += (y - dd) if c == get_upper_depth(v, dc)
    end
  }
  puts ans
}