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 = (n+1).times.map{[]}

pri = Array.new(n, 10)

m.times{
  l,r,d = gets.split.map &:to_i
  if l>r
    tmp = l
    l = r
    r = tmp
    d = -d
  end
  pri[l] = 2
  pri[r] = 2
  
  lrd[l] << [l,r,d]
}

hipos = {}

sel = 0
n.times{|i|
  e = pri[i+1]
  next if !e
  if 10 > e
    sel = i+1
    break
  end
}
q = [sel]
ch = Array.new(n+1, false)
ch[sel] = true

while q.size != 0
  now = q.shift
  lrd[now].each{|(l,r,d)|
    if  hipos[l] && hipos[r]
      #どちらもあるので正しいかみる
      if hipos[l]+d != hipos[r]
        puts 'No'
        exit
      end
    elsif hipos[l]
      #rがないのでいれる
      hipos[r] = hipos[l]+d
    elsif hipos[r]
      #lがないのでいれる
      hipos[l] = hipos[r]-d
    else
      #どちらもないのでいれる
      hipos[l] = 0
      hipos[r] = d
    end
    
    if !ch[r]
      q<<r
      ch[r] = true
    end
  }
end

max = 0
min = 10**9
n.times{|i|
  w = hipos[i+1]
  next if !w
  max = w if max < w
  min = w if min > w
}

if max-min <= (10**9)
  puts 'Yes'
else
  puts 'No'
end