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<cstdio>
#include<iostream>
using namespace std;
struct node{
	int p,num;
}a[100007];
int find(int x)
{
	if(a[x].p==x&&a[x].num==0){
		return x;
	}
	else 
	{
		int t=find(a[x].p);
		a[x].num=a[a[x].p].num+a[x].num;
		a[x].p=t;
		return t;
	}
}
int main()
{
	int n,m,i;
	cin>>n>>m;
	for(i=1;i<=n;i++){
		a[i].p=i;a[i].num=0;
	}
		
	for(i=1;i<=m;i++)
	{
		int l,r,x;
		scanf("%d%d%d",&l,&r,&x);
		int pl=find(l),pr=find(r);
		if(pl!=pr)
		{
			if(pl>pr)
			{
				a[pl].p=pr;
				a[pl].num=-(a[l].num-a[r].num+x);
			}
			else 
			{
				a[pr].p=pl;
				a[pr].num=a[l].num-a[r].num+x;
			}
		}
		else{
			if(a[r].num-a[l].num!=x)
			{
				cout<<"No"<<endl;
				return 0;
			}
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}