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)

class WeightedUnionFind
  def initialize(n)
    @parent = (0...n).to_a # 親の番号
    @rank = Array.new(n, 0) # 木の深さ
    @diff_weight = Array.new(n, 0) # 親との重みの差
  end

  def root(x)
    return x if @parent[x] == x
    r = root(@parent[x])
    @diff_weight[x] += @diff_weight[@parent[x]]
    @parent[x] = r
    return r
  end

  def weight(x)
    root(x)
    return @diff_weight[x]
  end

  def same?(x, y)
    root(x) == root(y)
  end

  def diff(x, y)
    # x と y とが同じグループにいるとき、weight(y) - weight(x) を返す
    # 同じグループでないならnil
    return nil if !same?(x, y)
    return weight(y) - weight(x)
  end

  def merge(x, y, w)
    # weight(y) - weight(x) = wとなるように結合
    w += weight(x)
    w -= weight(y)
    x, y = root(x), root(y)
    return false if x == y
    if @rank[x] < @rank[y]
      x, y = y, x
      w = -w
    end
    @rank[x] += 1 if @rank[x] == @rank[y]
    @parent[y] = x
    @diff_weight[y] = w
    return true
  end
end

N,M=gets.split.map(&:to_i)
uf= WeightedUnionFind.new(N)
M.times do
  l, r, d = gets.split.map(&:to_i)
  if uf.same?(l-1, r-1)
    if uf.diff(l-1, r-1) != d
      puts 'No'
      exit(0)
    end
  end
  uf.merge(l-1, r-1, d)
end

puts 'Yes'