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.chomp.split.map(&:to_i)
pos = {}
gr = {}
id = {}
idx = 0
m.times do
  l,r,d = gets.chomp.split.map(&:to_i)
  if !pos[l] && !pos[r]
    gr[idx] = [l,r]
    id[l],id[r] = idx,idx
    idx += 1
    pos[l] = 0
    pos[r] = d
  elsif !pos[l]
    gr[id[r]].push(l)
    id[l] = id[r]
    pos[l] = pos[r] - d
  elsif !pos[r]
    gr[id[l]].push(r)
    id[r] = id[l]
    pos[r] = pos[l] + d
  else
    if d != pos[r] - pos[l]
      if id[l] != id[r]
        diff = d - pos[r] + pos[l]
        gr[id[l]].concat(gr[id[r]])
        gr[id[r]].each do |k|
          pos[k] += diff
          id[k] = id[l]
        end
      else
        printf("No\n")
        exit
      end
    end
  end
end
printf("Yes\n")