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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
int n,m;
int x[100010];
queue<pair<int,int> >Q[100010];
queue<int> T;
int main(){
  cin>>n>>m;
  lol(i,m){
    int a,b,c;cin>>a>>b>>c;
    a--,b--;
    Q[a].push(make_pair(b,c));
    Q[b].push(make_pair(a,-c));
  }
  lol(i,n)x[i]=-mod;
  bool ans=true;
  lol(i,n){
    if(x[i]==-mod)T.push(i);
    else continue;
    x[i]=0;
    while(!T.empty()){
      int a=T.front();
      T.pop();
      while(!Q[a].empty()){
	int to=Q[a].front().first;
	int cost=Q[a].front().second;
	Q[a].pop();
	if(x[to]==-mod){
	  x[to]=x[a]+cost;
	  T.push(to);
	}
	else{
	  if(x[to]!=x[a]+cost)ans=false;
	}
      }
    }
  }
  cout<<(ans?"Yes":"No")<<endl;
  return 0;
}