Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

# https://www.leighhalliday.com/weighted-quick-union-find-algorithm-in-ruby
class UnionFind

  attr_accessor :nodes, :sizes

  def initialize(num)
    self.nodes = []
    self.sizes = []

    num.times do |n|
      self.nodes[n] = n
      self.sizes[n] = 1
    end
  end

  def root(i)
    # Loop up the chain until reaching root
    while nodes[i] != i do
      # path compression for future lookups
      nodes[i] = nodes[nodes[i]]
      i = nodes[i]
    end
    i
  end

  def union(i, j)
    rooti = root i
    rootj = root j

    # already connected
    return if rooti == rootj

    # root smaller to root of larger
    if sizes[i] < sizes[j]
      nodes[rooti] = rootj
      sizes[rootj] += sizes[rooti]
    else
      nodes[rootj] = rooti
      sizes[rooti] += sizes[rootj]
    end
  end

  def connected?(i, j)
    root(i) == root(j)
  end

end

N, M = gets.split.map &:to_i

uf = UnionFind.new(N+1)
coords = {}
lrd = $<.map{|s|
  l, r, d = s.split.map &:to_i
  uf.union(l, r)
  [l, r, d]
}

puts lrd.all?{|l, r, d|
  root = uf.root(l)
  coords[root] ||= {}
  if coords[root][l] && coords[root][r]
    coords[root][r] - coords[root][l] == d
  else
    if coords[root][l]
      coords[root][r] = coords[root][l] + d
    elsif coords[root][r]
      coords[root][l] = coords[root][r] - d
    else
      coords[root][l] = 0
      coords[root][r] = d
    end
    true
  end
} ? :Yes : :No