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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;
int n,m,poz[1000000+100];
bool obidjen[1000000+100];
struct node
{
    int l,r,d;
};
node a[1000000+100];
bool cmp(node a,node b)
{
    return min(a.r,a.l)<min(b.r,b.l);
}
int main()
{
    scanf("%d %d",&n,&m);
    for(int i = 1;i<=m;i++)
    {
        scanf("%d %d %d",&a[i].l,&a[i].r,&a[i].d);
    }
    sort(a+1,a+m+1,cmp);
  //  for (int i =1;i<=m;i++)printf("%d %d %d\n",a[i].r,a[i].l,a[i].d);

    for (int i = 1;i<=m;i++)
    {
        int idx = a[i].l;
        int idx2 = a[i].r;
        int smer = 1;
        if (a[i].r<a[i].l)idx = a[i].r,idx2 = a[i].l,smer = -1;

        if (!obidjen[idx])obidjen[idx]=1,poz[idx]=0;

        if (obidjen[idx2]&&poz[idx2]!=a[i].d*smer+poz[idx]){cout<<"NO"<<endl;return 0;}
        if (!obidjen[idx2])obidjen[idx2]=1,poz[idx2] = a[i].d*smer+poz[idx];

       // cout<<idx<<" "<<poz[idx]<<"   "<<idx2<<" "<<poz[idx2]<<endl;
    }
    cout<<"YES"<<endl;
    return 0;
}