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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var x,y,z,n,m:longint;
    i,t,w:longint;
    father,cost:array[0..1000000]of longint;
    flag:boolean;
function getfather(x:longint):longint;
var t:longint;
begin
  if father[x]=x then exit(x) else
  begin
    t:=getfather(father[x]);
    cost[x]:=cost[x]+cost[father[x]];
    father[x]:=t;
    getfather:=father[x];
  end;
end;
procedure join(x,y,z:longint);
var fx,fy:longint;
begin
  fx:=getfather(x);
  fy:=getfather(y);
  if fx<>fy then
  begin
    father[fx]:=fy;
    cost[fx]:=cost[y]-z-cost[x];
  end else
  if cost[y]-cost[x]<>z then flag:=false;
end;
begin
    flag:=true;
    read(n,m);
    for i:=0 to n do
    begin
      father[i]:=i;
      cost[i]:=0;
    end;
    for i:=1 to m do
    begin
      read(x,y,z);
      join(x,y,z);
    end;
    if flag then writeln('Yes') else writeln('No');
end.