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 main(void)
{
    long int n,m;
    scanf("%ld %ld",&n,&m);
    long int l[200005]={};
    long int r[200005]={};
    long int d[200005]={};
    for(int p=0;p<m;p++)scanf("%ld %ld %d",&l[p],&r[p],&d[p]);
    long int gr[100005]={0};
    long long int pl[100005]={};
    long int rer[100005]={114514};
    long int rel[100005]={114514};
    long int c=0;
    long long int x=0;
    for(int q=0;q<m;q++)
    {
        rer[l[q]]=r[q];
        rel[r[q]]=l[q];
        if(gr[l[q]]==gr[r[q]] && gr[r[q]]!=0)
        {
            if(pl[r[q]]-pl[l[q]]!=d[q])
            {
                printf("No");
                return 0;
            }
        }
        else if(gr[l[q]]!=0 && gr[r[q]]!=0)
        {
            gr[r[q]]=gr[l[q]];
            long int len=pl[l[q]]+d[q]-pl[r[q]];
            x=r[q];
            while(x!=114514)
            {
                gr[x]=gr[l[q]];
                pl[x]=pl[x]+len;
                x=rer[x];
            }
            x=r[q];
            while(x!=114514)
            {
                gr[x]=gr[l[q]];
                pl[x]=pl[x]+len;
                x=rel[x];
            }
        }
        else if(gr[l[q]]!=0)
        {
            gr[r[q]]=gr[l[q]];
            pl[r[q]]=pl[l[q]]+d[q];
        }
        else if(gr[r[q]]!=0)
        {
            gr[l[q]]=gr[r[q]];
            pl[l[q]]=pl[r[q]]-d[q];
        }
        else
        {
            pl[l[q]]=0;
            pl[r[q]]=d[q];
            c++;
            gr[l[q]]=c;
            gr[r[q]]=c;
        }
    }
    long long int max=pl[0];
    long long int min=pl[0];
    for(int ppp=0;ppp<n;ppp++)
    {
        if(max<pl[ppp])max=pl[ppp];
        if(min>pl[ppp])min=pl[ppp];
    }
    if(max-min>1000000000)
    {
        printf("No");
        return 0;
    }
    printf("Yes");
}