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;
typedef long long ll;
const int N=2e5+5;
ll n,m,fa[N],a[N];
int find(int x)
{
	if(x==fa[x])
		return x;
	int t=fa[x];
	int ft=find(t);
	a[x]+=a[t];
	return fa[x]=ft;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    	fa[i]=i,a[i]=0;
	ll u,v,d;
    bool ok=true;
    while(m--)
    {
    	cin>>u>>v>>d;
    	ll x=find(u),y=find(v);
    	if(x!=y)
    	{
    		if(x>y)
			{
				fa[x]=y;
				a[x]=a[v]-(a[u]+d);
			}		
			else
			{
				fa[y]=x;
				a[y]=a[u]-(a[v]-d);
			}
		}
		else 
		{
			ll t=a[v]-a[u];
			if(t!=d)
				ok=false;
		}
	}
	for(int i=1;i<=n;i++)
	{
		find(i);
		if(a[i]>1e9)
			ok=false;
	}
	puts(ok?"Yes":"No");
    
	    

    return 0;
}