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;

int N,M;
struct edge{ int from,to,cost; };

vector< edge > G[100000];

map<int,bool> visited;
map<int,int> d;

bool dfs(int pos,int sum){
  if(visited[pos]){
    if( d[pos] == sum )return true;
    else return false;
  }
  visited[pos]=true;
  d[pos]=sum;
  for(int i=0;i<(int)G[pos].size();i++){
    edge e=G[pos][i];
    if( !dfs(e.to, sum+e.cost) )return false;
  }
  return true;
}

int main(){
  cin>>N>>M;
  for(int i=0;i<M;i++){
    int L,R,D;
    cin>>L>>R>>D;
    L--,R--;
    G[L].push_back( (edge){L,R, D} );
    G[R].push_back( (edge){R,L,-D} );
  }

  bool ans=true;
  for(int i=0;i<N;i++){
    if(visited[i])continue;
    if( !dfs(i,0) ){
      ans=false;
      break;
    }
  }
  cout<< (ans?"Yes":"No") <<endl;
  return 0;
}