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 <iostream>
#include <algorithm>
#include <utility>
using namespace std;

#define INF 10000000000

int answer=1;
int checked[100000]={};

int L[200000], R[200000], D[200000];

pair<int, int> Li[200000]={}, Ri[200000]={};
int start_Li[100000]={};
int start_Ri[100000]={};

long long x[100000];

int check(int num){
	checked[num]=1;
	for(int i=start_Li[num]; num==Li[i].first; i++){
			
		if(x[ R[  Li[i].second ]]==INF){
			x[ R[  Li[i].second ]]=x[num]+D[Li[i].second];	
		}
		else if(x[ R[  Li[i].second ]]!=x[num]+D[Li[i].second]){
			answer=0;
		}
		
		if(checked[R[Li[i].second]]==0){
		check(R[Li[i].second]);
		}
	}
	for(int i=start_Ri[num]; num==Ri[i].first; i++){
			
		if(x[ L[ Ri[i].second  ] ]==INF){
			x[ L[ Ri[i].second  ] ]=x[num]-D[Ri[i].second];
		}else if(	x[ L[ Ri[i].second  ] ]!=x[num]-D[Ri[i].second]){
			answer=0;
		}
		
		if(checked[L[Ri[i].second]]==0){
		check(L[Ri[i].second]);

		}
	}
}

int main() {
	int N,M;
	cin >> N >> M;
	
	if(M==0){cout << "Yes"<<endl; return 0;}
	
	for(int i=0; i<M; i++){
		cin >> L[i] >> R[i] >> D[i];
		L[i]--;
		R[i]--;
	}
	

	for(int i=0; i<M; i++){
		Li[i].first=L[i];
		Li[i].second=i;
		Ri[i].first=R[i];
		Ri[i].second=i;
	}
	
	sort(Li,Li+M);
	sort(Ri,Ri+M);

	
	for(int i=1; i<N; i++){
		if(Li[i].first!=Li[i-1].first){
			start_Li[Li[i].first]=i;
		}
		if(Ri[i].first!=Ri[i-1].first){
			start_Ri[Ri[i].first]=i;
		}
	}
	

	for(int i=0; i<N; i++){
		x[i]=INF;
	}
	for(int i=0; i<N; i++){
		if(x[i]==INF){x[i]=0;}
		if(checked[i]==0){
		check(i);
		}
	}

	
	if(answer==0){
		cout<<"No"<<endl;
		return 0;
	}
	cout<<"Yes"<<endl;
	return 0;
}