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)

nm = gets.chomp.split(" ").map(&:to_i)
m = nm[1]

infos = []
m.times {
  infos.push( gets.chomp.split(" ").map(&:to_i) )
}

relations = {}
persons = []
infos.each {|info|
  l = info[0]
  r = info[1]
  d = info[2]

  if (l > r) then
    tmp = r
    r = l
    l = tmp
    d = -d
  end

  if persons.empty? then
    persons.push(l)
    persons.push(r)
  else
    current = l
    for i in 0..(persons.size-1) do
      if persons[i] < current then
        next
      end
      if current < persons[i] then
        persons.insert(i,current)
        if current == r then
          break
        else
          current = r
        end
      end
      if i == persons.size - 1 then
        if current == l then
          persons.push(l)
        end
        persons.push(r)
      end
    end
  end


  if relations.include?(l) then
    if relations[l].include?(r) then
      if relations[l][r] != d then
        puts "No"
        exit
      end
    else
      relations[l][r] = d
    end
  else
    relations[l] = {}
    relations[l][r] = d
  end

  for middle in 1..(persons.size-2) do
    for i in 0..(middle-1) do
      for j in (middle+1)..(persons.size-1) do
        m_person = persons[middle]
        i_person = persons[i]
        j_person = persons[j]
        if relations.include?(i_person) and relations[i_person].include?(m_person) and relations.include?(m_person) and relations[m_person].include?(j_person) then
          if relations[i_person].include?(j_person) and relations[i_person][j_person] != relations[i_person][m_person] + relations[m_person][j_person] then
            puts "No"
            exit
          end
          relations[i_person][j_person] = relations[i_person][m_person] + relations[m_person][j_person]
        end
      end
    end
  end
}

puts "Yes"