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 graph[100001];
int dist[100001];

int dfs(int i){
  if(graph[i] != i){
    int r = dfs(graph[i]);
    dist[i] += dist[graph[i]];
    return graph[i] = r;
  }else{
    return i;
  }
}

void init(int N){
  for(int i = 0;i < N;i++){
    graph[i] = i;
    dist[i] = 1;
  }
}

int main(void){
  int N,M;
  scanf("%d%d",&N,&M);

  init(N);
  int i,l,r,d;
  for(i = 0;i < M;i++){
    scanf("%d%d%d",&l,&r,&d);
    int g1 = dfs(l);
    int g2 = dfs(r);
    if(g1 != g2){
      graph[g2] = g1;
      dist[g2] = dist[l]+d-dist[r];
    }else{
      if(dist[r]-dist[l] != d){
        for(;i < M;i++) scanf("%d%d%d",&l,&r,&d);
        printf("No\n");
        return 0;
      }
    }
  }

  printf("Yes\n");
  return 0;
}