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>

int d[120000],f[120000],h[120000];
int i,m,n,x,y,dx,dy,fx,fy,z;

inline int getfather(int x)
{
	if (f[x]==x)
		return x;
	else
		return getfather(f[x]);
}

inline int getdis(int x)
{
	if (f[x]==x)
		return 0;
	else
		return d[x]+getdis(f[x]);
}

int main()
{
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;i++)
		f[i]=i,h[i]=1;
	for (i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		fx=getfather(x);
		fy=getfather(y);
		if (fx!=fy)
			if (h[fx]<h[fy])
				d[fx]=z-getdis(x)+getdis(y),f[fx]=fy,h[fy]=h[fy]+h[fx];
			else
				d[fy]=-z-getdis(y)+getdis(x),f[fy]=fx,h[fx]=h[fx]+h[fy];
		else
		{
			dx=getdis(x);
			dy=getdis(y);
			if (dx!=dy+z)
				return puts("No"),0;
		}
	}
	return puts("Yes"),0;
}