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: Time Limit Exceeded)

N, M = gets.split.map(&:to_i)
queries = M.times.map { gets.split.map(&:to_i) }

if M == 0
  puts 'Yes'
  exit 0
end

graph = Array.new(N) { [] }
indegree = Hash.new { 0 }
queries.each do |(l, r, d)|
  l -= 1
  r -= 1
  indegree[l] = 0
  indegree[r] += 1
  graph[l].push([r, d])
end

def has_cycle?(indegree, graph)
  queue = []
  indegree.each do |u, c|
    queue.push(u) if c == 0
  end
  count = queue.size
  while !queue.empty?
    u = queue.shift
    (graph[u]).each do |v|
      indegree[v] -= 1
      if indegree[v] == 0
        queue.push(v)
        count += 1
      end
    end
  end
  count == indegree.size
end

if has_cycle?(indegree, graph)
  puts 'No'
  exit 0
end

def bfs(graph, s)
  w_list = Array.new(graph.size, nil)
  w_list[s] = 0
  queue = [[s, nil, 0]]
  while !queue.empty?
    u, parent = queue.shift
    graph[u].each do |(v, w)|
      if w_list[v] && w_list[u] + w != w_list[v]
        return false
      elsif w_list[v].nil?
        w_list[v] = w_list[u] + w
        queue.push([v, w_list[v]])
      end
    end
  end
  true
end

indegree.each do |u, c|
  next unless c == 0
  unless bfs(graph, u)
    puts 'No'
    exit 0
  end
end
puts 'Yes'