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)

n, m = gets.split.map(&:to_i)
lrd  = []
m.times do
  l, r, d = gets.split.map(&:to_i)
  lrd << [l, d, r]
end

graph = {}
lrd.each do |l, d, r|
  graph[l] = {} if not graph.has_key?(l)
  graph[r] = {} if not graph.has_key?(r)
  graph[l][r] = d
  graph[r][l] = d
end

1.upto(n) do |k|
  1.upto(n) do |i|
    1.upto(n) do |j|
      if graph.has_key?(i) && graph.has_key?(j) && graph.has_key?(k)
        ij = graph[i][j] || Float::INFINITY
        ik = graph[i][k] || Float::INFINITY
        kj = graph[k][j] || Float::INFINITY
        if ij > ik + kj
          graph[i][j] = ik + kj
        end
      end
    end
  end
end

ok = true
lrd.each do |l, d, r|
  ok = false if graph[l][r] != d
  ok = false if graph[r][l] != d
end
puts ok ? "Yes" : "No"