Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
int main(){
  int n,m;
  cin>>n>>m;
  vector<P> v[111111];
  for(int i=0;i<m;i++){
    int l,r,d;
    cin>>l>>r>>d;
    v[l].push_back(P(r,d));
    v[r].push_back(P(l,-d));
  }
  int cos[111111],ans=1;
  fill(cos,cos+n+1,INT_MAX);
  for(int i=1;i<=n;i++){
    if(cos[i]==INT_MAX){
      queue<P> q;
      q.push(P(i,0));
      while(!q.empty()){
	int to=q.front().first;
	int cost=q.front().second;
	q.pop();
	if(cos[to]==INT_MAX){
	  cos[to]=cost;
	  for(int j=0;j<v[to].size();j++){
	    P p=v[to][j];
	    p.second+=cost;
	    q.push(p);
	  }
	}else if(cos[to]!=cost)ans=0;
      }
    }
  }
  if(ans)cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  return 0;
}