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: Runtime Error)

#include<stdio.h>

int main(){
  int d[200000],flag1=1,flag[100000],flagm[200000]={},flagr=1,flag2=0;
  long n,m,l[200000],r[200000],i,p[100000],g=2,max[50000],min[50000];
  scanf("%ld %ld",&n,&m);
  for(i=0;i<m;i++){
    scanf("%ld %ld %ld",&l[i],&r[i],&d[i]);
  }
  flag[1]=p[1]=max[1]=min[1]=1;
  do{
    flag2=0;
    flagr=1;
    for(i=0;i<m;i++){
      if(!flagm[i]){
        if(flag[r[i]]&&flag[l[i]]){
          flagm[i]=1;
          flag2=1;
          if(p[r[i]]-p[l[i]]!=d[i]){
            flag1=0;
            break;
          }
        }
        else if(flag[r[i]]){
          flagm[i]=1;
          flag[l[i]]=flag[r[i]];
          p[l[i]]=p[r[i]]-d[i];
          if(min[flag[l[i]]]>p[l[i]]) min[flag[l[i]]]=p[l[i]];
          flag2=1;
        }
        else if(flag[l[i]]){
          flagm[i]=1;
          flag[r[i]]=flag[l[i]];
          p[r[i]]=d[i]+p[l[i]];
          if(max[flag[l[i]]]<p[r[i]]) max[flag[l[i]]]=p[r[i]];
          flag2=1;
        }
      }
    }
    if(!flag2){
      for(i=0;i<m;i++){
        if(!flagm[i]){
          flagm[i]=flag[l[i]]=flag[r[i]]=g++;
          max[flag[l[i]]]=p[r[i]]=1;
          min[flag[l[i]]]=p[l[i]]=p[r[i]]-d[i];
        }
      }
    }
    for(i=0;i<m;i++){
      flagr*=flagm[i];
    }
  }while(!flagr&&flag1);
  for(i=1;i<g;i++){
    if(max[i]-min[i]>1000000000){
      flag1=0;
      break;
    }
  }
  if(flag1) printf("Yes\n");
  else printf("No\n");
}