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
	c,x:array[1..100000]of Longint;
	G1,G2:array[1..100000]of array of Longint;
	used:array[1..100000]of Boolean;
	n,m,i,l,r,d:Longint;
function dfs(u,p:Longint):Boolean;
var i:Longint;
begin
	used[u]:=True;
	for i:=0 to c[u]-1 do begin
		if G1[u,i]=p then continue;
		if used[G1[u,i]]and(x[G1[u,i]]<>x[u]+G2[u,i])then begin
			dfs:=False;
			exit;
		end;
		if not used[G1[u,i]]then begin
			x[G1[u,i]]:=x[u]+G2[u,i];
			if not dfs(G1[u,i],u)then begin
				dfs:=False;
				exit;
			end;
		end;
	end;
	dfs:=True;
end;
begin
	read(n,m);
	for i:=1 to m do begin
		read(l,r,d);
		inc(c[l]);inc(c[r]);
		setlength(G1[l],c[l]);
		setlength(G2[l],c[l]);
		setlength(G1[r],c[r]);
		setlength(G2[r],c[r]);
		G1[l,c[l]-1]:=r;
		G2[l,c[l]-1]:=d;
		G1[r,c[r]-1]:=l;
		G2[r,c[r]-1]:=-d;
	end;
	for i:=1 to n do begin
		if not used[i]then begin
			x[i]:=0;
			if not dfs(i,0)then begin
				writeln('No');
				exit;
			end;
		end;
	end;
	writeln('Yes');
end.