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 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i,a,b) for(int i=a;i<=b;i++)

typedef pair<int,int> ii;
typedef vector<ii> vii;

vector<vii> g;
int n,m,l,r,d;
int val[100001],ok;
bool visited[100001];

void dfs(int v) {
	visited[v]=true;
	rep(i,0,(int)g[v].size()-1) {
		ii p=g[v][i];
		if(visited[p.first]) {
			if(val[p.first]!=val[v]+p.second) {
					ok=false;
		return;
			}
		}
		else {
			val[p.first]=val[v]+p.second;
			dfs(p.first);
		}
		
	} 
}

signed main() {
	cin >> n >> m;
	g.assign(n+1,vii()) ;
	memset(visited,false,sizeof(visited));
	rep(i,1,m) {
		cin >> l >> r >> d;
		g[l].push_back(make_pair(r,d));
		g[r].push_back(make_pair(l,-d));
	}
	rep(i,1,n) {
		if(!visited[i]) {
			ok=true;
			dfs(i);
			if(ok==false) {
				cout << "no";
				return 0;
			}
		}
	}
	cout << "yes";
}