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)

def main()
  _n, m = gets.split.map { |t| t.to_i }
  if m == 0
    puts "Yes"
    exit
  end

  # person to loc
  p2l = {}
  m.times do
    loc_info = gets.split.map { |t| t.to_i }
    left, right, _ = loc_info
    if p2l[left]
      p2l[left].push(loc_info)
    else
      p2l[left] = [loc_info]
    end
    if p2l[right]
      p2l[right].push(loc_info)
    else
      p2l[right] = [loc_info]
    end
  end

  loc_hash = {}
  while p2l.size > 0
    if loc_hash.empty?
      person = p2l.keys[0]
    else
      # Check already appeared person first.
      for person in p2l.keys
        break if loc_hash[person]
      end
      person = p2l.keys[0] unless loc_hash[person]
    end
    list = p2l.delete(person)

    loc = loc_hash[person] || 0
    loc_hash[person] = loc
    for left, right, dist in list
      if left == person
        if loc_hash[right] and loc_hash[right] != loc + dist
          puts "No"
          exit
        end
        loc_hash[right] = loc + dist
      else
        if loc_hash[left] and loc_hash[left] != loc - dist
          puts "No"
          exit
        end
        loc_hash[left] = loc - dist
      end
    end
  end
  puts "Yes"
end

main if __FILE__ == $0