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)

class WeightUnionFindTree
  def initialize(n)
    @nodes = (0..n).map { |i| Node.new(i) }
  end

  def unite(a, b, w)
    pa = find(a)
    pb = find(b)
    return if pa == pb
    
    if @nodes[pa].rank < @nodes[pb].rank
      @nodes[pa].parent = pb
      @nodes[pa].weight = w - @nodes[a].weight + @nodes[b].weight
    else
      @nodes[pb].parent = pa
      @nodes[pb].weight = -w - @nodes[b].weight + @nodes[a].weight
      @nodes[pa].rank += 1 if @nodes[pa].rank == @nodes[pb].rank
    end
  end

  def same?(a, b)
    find(a) == find(b)
  end

  def find(a)
    return a if @nodes[a].parent == a
    @nodes[a].weight += @nodes[@nodes[a].parent].weight
    @nodes[a].parent = find(@nodes[a].parent)
  end
  
  def weight(a)
    find(a)
    @nodes[a].weight
  end
  
  def diff(a, b)
    weight(a) - weight(b)
  end
  
  def max_diff
    min = @nodes.map { |node| node.weight } .min
    max = @nodes.map { |node| node.weight } .max
    max - min
  end
  
  def parents
    @nodes.map(&:parent)
  end
  
  private 
  
  class Node
    attr_accessor :parent, :weight, :rank 
  
    def initialize(n)
      @parent = n
      @weight = 0
      @rank = 0
    end
  end
end

n, m = gets.chomp.split.map(&:to_i)
info = Array.new(m) { gets.chomp.split.map(&:to_i) }

tree = WeightUnionFindTree.new(n)
info.each do |l, r, d|
  if tree.same?(l, r)
    unless tree.diff(l, r) == d
      puts "No"
      exit
    end
  end
  tree.unite(l, r, d)
end
puts tree.max_diff <= 10 ** 9 ? "Yes" : "No"