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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

class Main {
static ArrayList<edge> []adj;
static boolean []visited;
static int []dist;
public static boolean dfs(int u,int d)
{
	visited[u]=true;
	boolean ans=true;
	if(d!=dist[u] && dist[u]!=-1)
		ans= false;
	
	dist[u]=d;
	for(edge e:adj[u])
	{
		if(visited[e.v])
		{
			ans&=dist[e.v]==d+e.w;
		}
		else
			ans &=dfs(e.v,d+e.w);
	}
	return ans;
}
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt(),m=sc.nextInt();
		adj=new ArrayList[n];
		visited=new boolean[n];
		dist=new int [n];
		Arrays.fill(dist, -1);
		for(int i=0;i<n;i++)
			adj[i]=new ArrayList();
		while(m-->0)
		{
			int u=sc.nextInt()-1,v=sc.nextInt()-1,w=sc.nextInt();
			adj[u].add(new edge(v,w));
			adj[v].add(new edge(u,-w));
			
		}
		boolean ans=true;
		for(int i=0;i<n;i++)
			if(!visited[i])
				ans&=dfs(i,0);
		System.out.println(ans?"Yes":"No");

	}
	static class edge
	{
		int v,w;
		edge(int a,int b)
		{
			v=a;w=b;
		}
	}
	static class Scanner
	{
		BufferedReader br;
		StringTokenizer st;
		public Scanner(InputStream S)
		{
			br=new BufferedReader(new InputStreamReader(S));
		}
		public String nextLine() throws IOException
		{
			return br.readLine();
		}
		public String next()  throws IOException
		{
			while(st==null || !st.hasMoreTokens())
				st=new StringTokenizer(nextLine());
			return st.nextToken();
		}
		public int nextInt()  throws IOException
		{
			return Integer.parseInt(next());
		}
		public double nextDouble()  throws IOException
		{
			return Double.parseDouble(next());
		}
		public boolean ready() throws IOException
		{
			return br.ready();
		}
	}
}