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: Accepted)

n,m = gets.chomp.split(" ").map(&:to_i)
p = Array.new(n,0)
v = Array.new(n+1,true)
adj = Array.new(n).map{Array.new}
edge = Array.new(m)
f = 0
v[0] = false

m.times do |i|
  l,r,d = gets.chomp.split(" ").map(&:to_i)
  edge[i] = [l-1,r-1,d]
  adj[l-1] << [r-1,d]
  adj[r-1] << [l-1,-d]
end

while f < n
  q = [f]
  while q.any?
    t = q.shift
    adj[t].each do |i,j|
      if v[i]
        q.push(i)
        v[i] = false
        p[i] = p[t]+j
      end
    end
  end
  for i in f..n
    if v[i]
      v[i] = false
      f = i
      break
    end
  end
end

ans = true

for i in 0..m-1
  if p[edge[i][1]]-p[edge[i][0]] != edge[i][2]
    ans = false
    break
  end
end

puts ans ? "Yes" : "No"