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)

neighbors_i = Array.new(N){[]}
neighbors_d = Array.new(N){[]}

M.times do |i|
    l, r, d = gets.chomp.split.map(&:to_i)
    l -= 1
    r -= 1
    #neighbors[l - 1].push([r - 1, d])
    #neighbors[r - 1].push([l - 1, -d])

    neighbors_i[l].push(r)
    neighbors_d[l].push(d)
    neighbors_i[r].push(l)
    neighbors_d[r].push(-d)
end

# diffs[j]: difference between i, j (i is smallest index of cluster)
# j is visited if diffs[j] is not nil
diffs = Array.new(N)

N.times do |i|
    next if diffs[i]

    stack = [i]
    diffs[i] = 0

    while stack.length > 0 do
        t = stack[-1]

        back = true
        #while neighbors[t].length > 0 do
        while neighbors_i[t].length > 0 do
            #ni, nd = neighbors[t].pop
            ni = neighbors_i[t].pop
            nd = neighbors_d[t].pop
            if diffs[ni]
                # cycle
                if diffs[t] - diffs[ni] + nd != 0
                    print 'No'
                    exit
                end
            else
                # not yet visited
                stack.push(ni)
                diffs[ni] = diffs[t] + nd
                back = false
                break
            end
        end

        stack.pop if back
    end
end

print 'Yes'