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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>

using namespace std;
int n,m;
int pa[233333],dis[233333];
pair<int,int> find(int x)
{
	if(pa[x]==0)return make_pair(x,0);
	else
	{
		pair<int,int> tmp=find(pa[x]);
		dis[x]+=tmp.second;pa[x]=tmp.first;
		return make_pair(pa[x],dis[x]);
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	int l,r,d;
	while(m--)
	{
		scanf("%d%d%d",&l,&r,&d);
		pair<int,int> pl=find(l),pr=find(r);
		if(pl.first==pr.first&&pl.second+d!=pr.second)
		{
			return printf("No\n"),0;
		}
		else if(pl.first!=pr.first)
		{
			pa[pr.first]=pl.first;
			dis[pr.first]=d-pr.second+pl.second;
		}
	}
	printf("Yes\n");
	return 0;
}