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 <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cmath>
using namespace std;
const int N=100005;
int n,m,fa[N],d[N];
int getf(int k){
	if (fa[k]==k)
		return k;
	int f=fa[k];
	fa[k]=getf(fa[k]);
	d[k]+=d[f];
	return fa[k];
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		fa[i]=i;
	memset(d,0,sizeof d);
	for (int i=1;i<=m;i++){
		int L,R,D;
		scanf("%d%d%d",&L,&R,&D);
		if (getf(L)==getf(R)){
			if (d[L]-d[R]!=D){
				puts("No");
				return 0;
			}
		}
		else {
			int A=getf(L),B=getf(R);
			fa[B]=A,d[B]=d[L]-D-d[R];
		}
	}
	puts("Yes");
	return 0;
}