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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define debug(x) cout<<#x<<": "<<x<<endl
#define DEBUG(x) cout<<#x<<": "<<x<<endl
int powint(int a, int x){
	int res = 1;
	REP(i,x){
		res = res *a; 
	}
	return res;
}

using namespace std;

class Point {
public:
	int jiku;
	long long int zahyou;
public:
	Point(){
		jiku=-1;
		zahyou=0;
	}
};

class Bar{
public:
	set<int> points;
};

int main(){
	int N,M;
	cin>>N>>M;
	vector<Point> x(N);
	vector<Bar> b(N);
	bool dame=false;
	REP(i,M){
		int L,R,D;
		cin>>L>>R>>D;
		L--;
		R--;
		if(x[L].jiku==-1 && x[R].jiku==-1){
			x[L].jiku=L;
			x[R].jiku=L;
			x[L].zahyou=0;
			x[R].zahyou=D;
			b[L].points.insert(L);
			b[L].points.insert(R);
		} else if (x[L].jiku!=-1 && x[R].jiku!=-1){
			if(x[L].jiku == x[R].jiku){
				if(x[R].zahyou-x[L].zahyou!=D){
					dame=true;
					break;
				}
			} else {
				//x[L].jikuのほうにマージする
				int prejiku = x[R].jiku;
				long long int prezahyou = x[R].zahyou;
				x[R].jiku = x[L].jiku;
				x[R].zahyou = x[L].zahyou + D;
				for( auto p: b[prejiku].points){
					if(p==R) continue;
					//debug(p);
					x[p].jiku = x[L].jiku;
					x[p].zahyou = x[R].zahyou+x[p].zahyou-prezahyou;//なりたい座標
					b[x[L].jiku].points.insert(p);
				}
				b[x[L].jiku].points.insert(R);
			}
		} else if(x[L].jiku!=-1){
			x[R].jiku=x[L].jiku;
			x[R].zahyou=x[L].zahyou+D;
			b[x[R].jiku].points.insert(R);
		} else {
			x[L].jiku=x[R].jiku;
			x[L].zahyou=x[R].zahyou-D;
			b[x[L].jiku].points.insert(L);
		}
		/*
		REP(i,N){
			cout<<"Bar"<<i+1<<": ";
			for(auto p: b[i].points){
				cout<<p+1<<"("<<x[p].zahyou<<") ";
			}
			cout<<endl;
		}
		cout<<endl;
		*/
	}
	if(dame){
		cout<<"No"<<endl;
	} else {
		cout<<"Yes"<<endl;
	}
	return 0;
}