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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.List;
import java.util.Scanner;
import java.util.TreeMap;

class Edge{
	int a,b,c;
	public Edge(int A,int B,int C){
		a=A;
		b=B;
		c=C;
	}
}
class Main{
	static Scanner s=new Scanner(System.in);

	static int gInt(){
		return Integer.parseInt(s.next());
	}

	public static void main(String[]$){
		final long INIT=Long.MAX_VALUE;

		TreeMap<Integer,List<Edge>>a=new TreeMap<>();
		int n=gInt(),m=gInt();
		long[] pos=new long[n];
		Arrays.fill(pos,INIT);

		for(int q=0;q<m;++q){
			int l=gInt()-1,r=gInt()-1,d=gInt();
			a.computeIfAbsent(l,(i)->new ArrayList<>(1)).add(new Edge(l,r,d));
			a.computeIfAbsent(r,(i)->new ArrayList<>(1)).add(new Edge(r,l,-d));
		}

		Deque<Edge> d=new ArrayDeque<>();
		while(!a.isEmpty()){
			d.addAll(a.pollFirstEntry().getValue());
			while(!d.isEmpty()){
				Edge o=d.poll();
				if(pos[o.a]==INIT){
					pos[o.a]=0;
				}
				if(pos[o.b]==INIT){
					pos[o.b]=pos[o.a]+o.c;
					List<Edge> v=a.remove(o.b);
					if(v!=null)
						d.addAll(v);
				}
				if(pos[o.b]-pos[o.a]!=o.c){
					System.out.println("No");
					return;
				}
			}
		}
		System.out.println("Yes");
	}
}