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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int N=400005;
int cnt,h[N],to[N],las[N],c[N];
int vis[N],val[N],n,m,u,v,w;

void dfs(int x)
{
	vis[x]=1;
	int i;
	for(i=h[x];i;i=las[i])
	    if(!vis[to[i]])
	    {
	    	val[to[i]]=val[x]+c[i];
	    	dfs(to[i]);
	    }
	    else if(val[to[i]]!=val[x]+c[i])
	    {
	    	printf("No");
	    	exit(0);
	    }
}

int main()
{
	scanf("%d%d",&n,&m);
	int i;
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		las[++cnt]=h[u],h[u]=cnt,to[cnt]=v,c[cnt]=w;
		las[++cnt]=h[v],h[v]=cnt,to[cnt]=u,c[cnt]=-w;
	}
	for(i=1;i<=n;i++)
	    if(!vis[i])
	        dfs(i);
	printf("Yes");
	return 0;
}