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 <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int N, M;
int p[100005];
ll dist[100005];
int find(int a){
	if(a==p[a]) return a;
	int pa=p[a];
	p[a]=find(pa);
	dist[a]+=dist[pa];
	return p[a];
}
bool merge(int a,int b,int d){
	int pa=find(a), pb=find(b);
	if(pa==pb){
		return dist[a]==dist[b]+d;
	}
	else{
		p[pa]=pb;
		dist[pa]=d-dist[a]+dist[b];
		return 1;
	}
}
int main(){
	scanf("%d %d",&N,&M);
	for(int i=1;i<=N;i++)
		p[i]=i;
	while(M--){
		int L,R,D; scanf("%d %d %d",&L,&R,&D);
		if(!merge(L,R,D)) return !printf("No");
	}
	puts("Yes");

}