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;
void dfs(vector<vector<vector<int>>> &g,vector<int> &p,bool &f,vector<bool> &c,int x){
  if(!f)
    return;
  c.at(x)=1;
  for(auto v:g.at(x)){
    int y=v.at(0);
    int d=v.at(1);
    if(c.at(y)){
      if(p.at(y)!=p.at(x)+d)
        f=0;
      continue;
    }
    p.at(y)=p.at(x)+d;
    dfs(g,p,f,c,y);
  }
  return;
}
int main(){
  int n,m;
  cin>>n>>m;
  vector<vector<vector<int>>> g(n);
  for(int i=0;i<m;i++){
    int l,r,d;
    cin>>l>>r>>d;
    l--;
    r--;
    g.at(l).push_back({r,d});
    g.at(r).push_back({l,-d});
  }
  vector<int> p(n,0);
  bool f=1;
  vector<bool> c(n,0);
  for(int i=0;i<n;i++){
    if(c.at(i))
      continue;
    dfs(g,p,f,c,i);
  }
  cout<<(f?"Yes":"No")<<endl;
}