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<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=100005;

int n,m;
int fa[N],dis[N];
int find(int x) 
{
	if (x==fa[x]) return x;
	int t=find(fa[x]);
	dis[x]+=dis[fa[x]];
	return fa[x]=t;
}


int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) fa[i]=i;
	
	int x,y,d,fx,fy;
	bool o=true;
	while (m--) 
	{
		scanf("%d%d%d",&y,&x,&d);
		if (!o) continue;

		fx=find(x),fy=find(y);
		if (fx==fy) 
		{
			if (dis[x]-dis[y]!=d) o=false;
		}
		else 
		{
			fa[fx]=fy;
			dis[fx]=-dis[x]+dis[y]+d;
		}
	}
	if (o) puts("Yes");else puts("No");
	return 0;
}