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: Accepted)

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

typedef long long ll;

const int NONDEF=-1,NMAX=1e5;
int n,m,nbGauche[NMAX],nbDroite[NMAX],distances[NMAX];
vector< pair<int,int> > voisins[NMAX];
queue< pair<int,int> > enCours;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    cin >> n >> m;
    int l,r,d;
    for(int i=0;i<m;i++){
		cin >> l >> r >> d;
		l--;r--;
		voisins[l].push_back( make_pair(r,d) );
		nbGauche[l]++;
		nbDroite[r]++;
	}
	
	for(int i=0;i<n;i++){
		distances[i]=NONDEF;
		if(nbDroite[i]==0){
			distances[i]=0;
			enCours.push(make_pair(i,0));
		}
	}
	if(enCours.empty()){
		cout << "No\n";
		return 0;
	}
	int sz,actuel,distance,voisin,p;
	
	while(!enCours.empty()){
		actuel=enCours.front().first;
		distance=enCours.front().second;
		enCours.pop();
		sz=voisins[actuel].size();
		
		for(int i=0;i<sz;i++){
			voisin=voisins[actuel][i].first;
			p=voisins[actuel][i].second;
			
			if(distances[voisin]!=NONDEF && distance+p!=distances[voisin]){
				cout << "No\n";
				return 0;
			}
			distances[voisin]=distance+p;
			nbDroite[voisin]--;
			if(nbDroite[voisin]==0){
				enCours.push(make_pair(voisin,distances[voisin]));
			}
		}
	}
	
	for(int i=0;i<n;i++){
		if(distances[i]==NONDEF && (nbGauche[i]!=0 || nbDroite[i]!=0)){
			cout << "No\n";
			return 0;
		}
	}
	
	cout << "Yes\n";
			
			
    return 0;
}