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)

MOD=10**9+7
cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def desc(ar) ar.sort!{|x,y|y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+) end

def bs(ar,eq,n) en=(eq==1 ? :>= : :>); ar.bsearch_index{|x|x.send(en,n)} end
def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end

def puts_yesno(can) puts(can ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def debug(k,v) puts "#{k}:#{v}" end
########### ( ˘ω˘ )ｽﾔｧ… ###########
n, m = gsmi
lrd =
  m.times.map do
    ar = gsmi
    ar = [ar[1],ar[0],-ar[2]] if ar[0]>ar[1]
    ar
  end.sort

x = Array.new(n)

lrd.each do |l, r, d|
  if x[l-1].nil? && x[r-1].nil?
    x[l-1] = 0
    x[r-1] = x[l-1] + d
  elsif x[l-1].nil?
    x[l-1] = x[r-1] - d
  elsif x[r-1].nil?
    x[r-1] = x[l-1] + d
  elsif x[r-1] != x[l-1] + d
    putsend('No')
  end
end

puts 'Yes'