Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int n,m,d[100010],ta[200010],nt[400010],to[400010],co[400010],q[100010],u[100010]={};
int main(){
	int a,b,c,r,t,i,j;
	scanf("%d %d",&n,&m);
	for(i=0;i<n;i++)ta[i+1]=-1;
	for(i=0;i<m;i++){
		scanf("%d %d %d",&a,&b,&c);
		nt[i  ]=ta[to[i+m]=a];
		nt[i+m]=ta[to[i  ]=b];
		co[ta[a]=i  ]= c;
		co[ta[b]=i+m]=-c;
	}
	for(i=1;i<=n;i++){
		if(u[i])continue;
		u[q[t=0]=i]=1;
		d[i]=0;
		for(r=1;r-t;t++){//printf(":%d %d\n",q[t],d[q[t]]);
			for(j=ta[q[t]];j+1;j=nt[j]){//printf("%d %d %d\n",q[t],to[j],co[j]);
				if(u[to[j]]){
					if(d[to[j]]-d[q[t]]-co[j])goto END;
				}
				else u[q[r++]=to[j]]=1;
				d[to[j]]=d[q[t]]+co[j];
			}
		}
	}
	printf("Yes\n");
	return 0;
	END:;
	//for(i=0;i<n;i++)printf("%d ",d[i+1]);printf("\n");
	printf("No\n");
	return 0;
}