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>
#include <stdlib.h>
int g[100001];
int di[100001];

int dfs(int i){
    if(g[i]!=i){
        int r = dfs(g[i]);
        di[i]+=di[g[i]];
        //printf("di[%d]=%d\n",i,di[i]);
        return r;
    }else{
        return i;
    }
}
int main(){
    int i,j;
    int N,M;
    scanf("%d %d",&N,&M);
    for(i=1;i<=N;i++)g[i]=i;
    for(i=0;i<M;i++){
        int l,r,d;
        scanf("%d %d %d",&l,&r,&d);
        int g1=dfs(l);
        int g2=dfs(r);
        if(g1!=g2){
            g[g2]=l;
            di[g2]=d-di[r];
        }else{
            //printf("di[%d](%d)-di[%d](%d)!=%d\n",r,di[r],l,di[l],d);
            if(di[r]-di[l]!=d){
                printf("No\n");
                return 0;
            }
        }
    }
    //printf("%d %d %d\n",di[1],di[2],di[3]);
    printf("Yes\n");
}