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: Wrong Answer)

#include<stdio.h>
int ta[100010],to[400040],co[400040],nt[400040],d[100010];
int main(){
  int n,m,i,j,t,r,a,b,c,q[100010],u[100010]={};
  for(i=0;i<100005;i++)d[i]=ta[i]=-1;
  scanf("%d %d",&n,&m);
  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[b]=i+m]=-c;
    co[ta[a]=i  ]= c;
  }
  for(j=1;j<=n;j++){
    if(u[j])continue;
    d[j]=1e9;
    q[t=0]=j;
    for(r=1;r-t;t++){
      u[q[t]]=1;
      //for(i=0;i<n;i++)printf("%d ",d[i+1]);printf("\n");
      for(i=ta[q[t]];i+1;i=nt[i]){printf("%d %d\n",q[t],to[i]);
	if(d[to[i]]>0&&d[to[i]]!=d[q[t]]+co[i])goto END;
	if(u[to[i]])continue;
	d[to[i]]=d[q[t]]+co[i];
	q[r++]=to[i];
      }
    }
  }
  printf("Yes\n");
  return 0;
 END:;
  printf("No\n");
  return 0;
}