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: Wrong Answer)

var
    n,m,i,j: Longint;
    L,R,D: array[1..200000] of Longint;
    position:array[1..100000] of Longint;
    flag:Boolean;
begin
    flag:=true;
    readln(n,m);
    for i := 1 to n do
    begin
        position[i]:=-2147483648;
    end;
    for i := 1 to m do
    begin
        readln(L[i],R[i],D[i]);
        if (position[L[i]]=-2147483648)and(position[R[i]]=-2147483648) then 
        begin
            position[L[i]]:=0;
            position[R[i]]:=D[i];
        end
        else 
        begin
            if (position[L[i]]<>-2147483648) and (position[R[i]]<>-2147483648) then 
            begin
                if position[L[i]]+D[i]<>position[R[i]] then
                begin
                    flag:=false;
                    break;
                end;
            end
            else
            begin
                if position[L[i]]<>-2147483648 then
                begin
                    position[R[i]]:=position[L[i]]+D[i];
                    for j := 1 to i-1 do
                    begin
                        if L[j]=R[i] then 
                        begin
                            position[R[j]]:=position[L[j]]+D[j];
                        end
                        else if R[j]=R[i] then
                        begin
                            position[L[j]]:=position[R[j]]-D[j];
                        end;
                    end;
                end
                else
                begin
                    position[L[i]]:=position[R[i]]-D[i];
                    for j := 1 to i-1 do
                    begin
                        if L[j]=L[i] then 
                        begin
                            position[R[j]]:=position[L[j]]+D[j];
                        end
                        else if R[j]=L[i] then
                        begin
                            position[L[j]]:=position[R[j]]-D[j];
                        end;
                    end;
                end;
            end;
        end;
    end;
    if flag then 
    begin
        writeln('Yes');
    end
    else 
    begin
        writeln('No');
    end;
end.