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.ArrayDeque;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Scanner;

/**
 * https://arc090.contest.atcoder.jp/tasks/arc090_b
 */
public class Main {

	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Node[] nodes = new Node[N];
		for(int i=0; i<N; i++) nodes[i] = new Node(i);
		int M = sc.nextInt();
		for(int i=0; i<M; i++){
			int l = Integer.parseInt(sc.next())-1;
			int r = Integer.parseInt(sc.next())-1;
			int d = Integer.parseInt(sc.next());
			nodes[l].neighbor.put(nodes[r], d);
		}
		sc.close();
		List<Node> list = topologicalSort(nodes);
		
		String ans = "Yes";
		if(list.size() == N){
			for(Node n:list){
				if(n.x==-1){
					if(!dfs(n,0)){
						ans = "No";
						break;
					}		
				}
			}
		}else{
			ans = "No";
		}
		
		System.out.println(ans);
		
	}
	
	static boolean dfs(Node node, long current){
		node.x = current;
		for(Entry<Node, Integer> n:node.neighbor.entrySet()){
			Node nn = n.getKey();
			int d = n.getValue();
			if(nn.x!=-1){
				if(nn.x != current+d) return false;
			}else{
				if(!dfs(nn, current+d)){
					return false;
				}	
			}
		}
		return true;
	}
	
	static class Node{
		int id;
		long x = -1;
		Map<Node,Integer> neighbor = new HashMap<>();
		int color = -1;
		int status = -1;
		int inDegree = 0;
		Node(int id){
			this.id = id;
		}
	}
	
	static List<Node> topologicalSort(Node[] nodeList){
		List<Node> sortedList = new ArrayList<>();
		for(Node node:nodeList)
			for(Node n:node.neighbor.keySet())
				n.inDegree++;
		for(Node node:nodeList){
			if(node.status!=-1 || node.inDegree>0) continue;
			Queue<Node> que = new ArrayDeque<>();
			node.status=1;
			que.add(node);
			while(!que.isEmpty()){
				Node current = que.remove();
				sortedList.add(current);
				for(Node n:current.neighbor.keySet()){
					if(--n.inDegree==0 && n.status==-1){
						n.status=1;
						que.add(n);
					}
				}
			}
		}
		return sortedList;
	}

}