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>
using namespace std;
typedef pair<int,int> P;

int n, m, ans=1;

vector<P> G[100005];
int used[100005];
map<int,int> A;
int minx, maxx;

void dfs(int x){
  
  if(used[x]) return ;
  used[x]=1;
  
  for(int i=0;i<G[x].size();i++){
    int nx=G[x][i].first;
    int d=G[x][i].second;
    if(A.count(nx)&&A[nx]!=A[x]+d){
      ans=0;
      break;
    }
    A[nx]=A[x]+d;
  }
  
  for(int i=0;i<G[x].size();i++){
    int nx=G[x][i].first;
    dfs(nx);
  }
  
}

int main(){
  
  cin>>n>>m;
  
  for(int i=0;i<m;i++){
    
    int l, r, d;
    cin>>l>>r>>d;
    
    G[l].push_back(P(r,d));
    G[r].push_back(P(l,-d));
  }
  
  for(int i=1;i<=n;i++){
    if(!used[i]){
      minx=0;
      maxx=0;
      A[i]=0;
      dfs(i);
      if(maxx-minx>1e9+1){
	ans=0;
	break;
      }
    }
  }
  
  if(ans) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  
  return 0;
}