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=1e5+10;
int n,m,x,y,z,fa[N],sum[N];
int Read()
{
	int x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return f*x;
}
int Get(int x)
{
	if(fa[x]==x) return x;
	int f=fa[x];
	fa[x]=Get(f);
	sum[x]+=sum[f];
	return fa[x];
}
void merge(int a,int b,int c)
{
	int f1=Get(a),f2=Get(b);
	fa[f1]=f2;
	sum[f1]=-sum[a]+sum[b]+c;
}
int main()
{
//	freopen("line.in","r",stdin);
	//freopen("line.out","w",stdout);
	n=Read(),m=Read();
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		x=Read(),y=Read(),z=Read();
		int f1=Get(x),f2=Get(y);
		if(f1==f2)
		{
			if(sum[x]-sum[y]!=z)
			{
				cout<<"No";
				return 0;
			}
		}
		else
		merge(x,y,z);
	}	
	cout<<"Yes";
	return 0;
}