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 UnionFind_Potential

implicit none
integer,allocatable,dimension(:)::par
!parentより
integer,allocatable,dimension(:)::diff_weight

contains

recursive function root(x) result(res)
  integer :: x,res

  if(par(x) /= x)then
    diff_weight(x)=diff_weight(x)+diff_weight(par(x))
    par(x) = root(par(x))
  endif
  res = par(x)
end function

subroutine unite(x,y,w)
  !x+w=yとなるように繋げます
  integer :: x,y,w,rx,ry,rw
  !rx,ryはそれぞれの根っこ
  !rwはrxとryの差分です
  
  rx = root(x);ry = root(y)
  
  !rx + weight(x) +w=ry+weight(y)から
  rw = w + weight(x)-weight(y)
  
  if(rx /= ry)then
    par(ry) = rx
    diff_weight(ry)=rw
  endif
end subroutine

function weight(x)
  integer::x,weight
  if(root(x)==x)then
    weight=0
  else
    weight=diff_weight(x)
  endif
end function

function diff(x,y)
  integer::x,y,diff
  diff=weight(y)-weight(x)
end function
end module UnionFind_Potential

use UnionFind_Potential

integer n,m
integer l,r,d
read*,n,m
allocate(par(n),diff_weight(n))

!初期化
do i=1,n
  par(i)=i
end do
diff_weight=0

do i=1,m
  read*,l,r,d
  if(root(l)/=root(r))then
    call unite(l,r,d)
  else
    if(diff(l,r)/=d)then
      print"(A)","No"
      STOP
    end if
  endif
end do

print"(A)","Yes"

end