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>
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 100000
#define MM 200000
struct edge{int nx,t,w;}e[MM*2+5];
int h[MN+5],en,u[MN+5],z[MN+5],O;
inline void ins(int x,int y,int w)
{
	e[++en]=(edge){h[x],y,w};h[x]=en;
	e[++en]=(edge){h[y],x,-w};h[y]=en;
}
void dfs(int x)
{
	u[x]=1;
	for(int i=h[x];i;i=e[i].nx)
		if(!u[e[i].t])z[e[i].t]=z[x]+e[i].w,dfs(e[i].t);
		else O|=z[e[i].t]!=z[x]+e[i].w;
}
int main()
{
	int n,m,x,y,i;
	n=read();m=read();
	while(m--)x=read(),y=read(),ins(x,y,read());
	for(i=1;i<=n;++i)if(!u[i])dfs(i);
	puts(O?"No":"Yes");
}