Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in JAVA):

Solution:
import java.util.*;

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");
	}
}