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 long long LL;
LL dst[100010]={0};
LL fat[100010];
LL n,m,l,r,d;
int getfat(int x){
	if(x==fat[x])return x;
	LL i=getfat(fat[x]);
	dst[x]+=dst[fat[x]];
	fat[x]=i;
	return i;
}
int main(){
	LL i,j,k;
	for(i=0;i<100010;i++)fat[i]=i;
	cin>>n>>m;
	for(i=0;i<m;i++){
		cin>>l>>r>>d;
		j=getfat(l);
		k=getfat(r);
		if(j==k&&d!=dst[l]-dst[r]){
			cout<<"No"<<endl;
			return 0;
		}
		if(j==k&&d==dst[l]-dst[r])	continue;
		if(j!=k){
			if(d>dst[l]){
				fat[j]=k;
				dst[j]=dst[r]+d-dst[l];
			}
			else if(d+dst[r]>dst[l]){
				fat[j]=k;
				dst[j]=dst[r]+d-dst[l];
			}
			else if(d+dst[r]<dst[l]){
				fat[k]=j;
				dst[k]=dst[l]-d-dst[r];
			}
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}