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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

module Weighted_UF
    
    integer, allocatable, dimension(:) :: par, rank, diff_weight

    contains
    
    subroutine init(N)
        integer :: N
        allocate(par(N), rank(N), diff_weight(N))
        do i = 1, N
            par(i) = i
            rank(i) = 0
            diff_weight(i) = 0
        end do
    end subroutine

    recursive function root(x) result (res)
        integer :: x, res
        if (par(x) == x)then 
            res = x
        else
            res = root(par(x))
            diff_weight(x) = diff_weight(x) + diff_weight(par(x))
            par(x) = res
        end if
    end function

    function weight(x) result (res)
        integer :: x, res, r
        res = diff_weight(x)
        r = root(x)
    end function

    function same(x, y) result (res)
        integer :: x, y
        logical :: res
        res = (root(x) == root(y))
    end function
    
    subroutine unite(x, y, w)
        integer :: x, y, w
        w = w + weight(x) - weight(y)
        x = root(x); y = root(y)
        if (x /= y) then
            if(rank(x) < rank(y)) then
                x = xor(x, y); y = xor(y, x); x = xor(x, y) !swap
                w = -w
            end if
            if(rank(x) == rank(y)) then
                rank(x) = rank(x) + 1
            end if
            par(y) = x
            diff_weight(y) = w
        end if
    end subroutine

    function diff(x, y) result (res)
        integer :: x, y, res
        res = weight(y) - weight(x)
    end function
    
    subroutine print_UF(N)
        integer :: N
        print"(a)", parents
        do i = 1, N
            print*, "Number :", i , " " , par(i)
        end do
    end subroutine
end module Weighted_UF

program main
    use Weighted_UF

    integer N, M, l, r, d
    read*,N, M

    call init(N)

    do i = 1, M
        read*, l, r, d
        if (same(l, r)) then
            if (diff(l, r) /= d) then
                print*, l, r
                print"(a)", "No"
                stop
            end if
        else
            call unite(l, r, d)
        end if
        
    end do

    print"(a)", "Yes"
end