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)
ls = []
rs = []
ds = []
m.times do
  l, r, d = gets.chomp.split(' ').map(&:to_i)
  ls << l
  rs << r
  ds << d
end

def find(l, r, dists, used)
  return dists[l][r] if dists.fetch(l, {})[r]
  return nil if used.fetch(l, {})[r]
  set(used, l, r, true)

  unless dists.key?(l)
    dists[l] = {}
    return nil
  end

  keys = dists[l].keys
  keys.each do |to|
    dist = dists[l][to]
    v = find(to, r, dists, used)
    next if v.nil?

    dists[l][r] = dist + v
    return dists[l][r]
  end
  return nil
end

def set(hash, parent, key, value)
  if hash.key?(parent)
    hash[parent][key] = value
  else
    hash[parent] = { key => value }
  end
end

dists = {}
0.upto(m - 1).each do |i|
  l = ls[i]
  r = rs[i]
  d = ds[i]
  dist = find(l, r, dists, {})
  if dist.nil?
    set(dists, l, r, d)
    set(dists, r, l, -1 * d)
  elsif dist != d
    puts 'No'
    exit 0
  end
end
puts 'Yes'