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 <bits/stdc++.h> 
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#define each(it,o) for(auto it= (o).begin(); it != (o).end(); ++ it)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
#define inrep(t) int t;cin>>t; while(t--)
#define pb push_back
#define mp make_pair 
using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi; 
typedef pair<int,int> pii;
typedef vector<pii > vpii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef vector<pll > vpll; 
typedef vector<string> vs;
typedef long double ld;
const int up= int(1e5)+5; 
vpii adj[up];
bool visited[up];   
int n,m;
int root; 
bool pos=1; 
vll x(up,-INFL); 
void dfs(int u, int p){
	//cerr<<u<<" "<<p<<endl; 
	if(pos==0) return; 
	visited[u]=1;
	reu(i,0,adj[u].size()){
		int v=adj[u][i].first; 
		ll d=adj[u][i].second; 
		if(v==p) continue; 
		
		if(x[v]==-INFL) { x[v]=x[u]+d; }
		else if((x[v]-x[u])!=d){
			pos=0; return; 
		}
		if(!visited[v]){
		dfs(v,u); 
		}
	} 

}
int main(){
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL); 
	cout.tie(NULL); 
	cin>>n>>m; 
	reu(i,0,m){
		int u,v,d; cin>>u>>v>>d; 
		adj[u].pb(mp(v,d));
		adj[v].pb(mp(u,-d));  
	}
	reu(i,1,n+1) {
		if(!visited[i]){
			x[i]=0; 
			dfs(i,0); 
		}
		if(!pos) break; 
	}
	if(pos) cout<<"Yes"; 
	else cout<<"No"; 

}