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;
struct node{
	int f,dis;
};
node f[100001];
int findx(int x){
	if(f[x].f==x&&f[x].dis==0)return x;
	int k=findx(f[x].f);
	f[x].dis=f[f[x].f].dis+f[x].dis;
	f[x].f=k;
	return k;
}
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		f[i].f=i;
		f[i].dis=0;
	}
	for(int i=1;i<=m;i++){
		int a,b,d;
		cin>>a>>b>>d;
		int k1=findx(a),k2=findx(b);
		if(k1!=k2){
			/*if(a>b){
				swap(a,b);
				d=-d;
			}*/
			if(k1>k2){
				f[k1].f=k2;
				f[k1].dis=-(f[a].dis-f[b].dis+d);
			}
			else{
				f[k2].f=k1;
				f[k2].dis=f[a].dis-f[b].dis+d;
			}
		}
		else{
			if(f[b].dis-f[a].dis!=d){
				cout<<"No";
				return 0;
			}
		}
	}
	cout<<"Yes";
}