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 bian{
	int x,to,next;
}edge[400001];
int cnt,head[100001],dis[100001];
bool vis[100001],f;
 
inline void add(int u,int v,int x){
	edge[++cnt].x=x,edge[cnt].to=v,edge[cnt].next=head[u],head[u]=cnt;
}
 
inline void dfs(int x){
	for(int i=head[x];i;i=edge[i].next){
		if(!vis[edge[i].to]){
			vis[edge[i].to]=1;
			dis[edge[i].to]=dis[x]+edge[i].x;
			dfs(edge[i].to);
		}
		else if(dis[edge[i].to]!=dis[x]+edge[i].x){
			f=0;
			return;
		}
	}
}
 
int main()
{
	ios::sync_with_stdio(false);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int l,r,d;
		cin>>l>>r>>d;
		add(l,r,-d),add(r,l,d);
	}
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			vis[i]=1;
			f=1;
			dfs(i);
			if(!f||dis[i]!=0){
				cout<<"No";
				return 0;
			}
		}
	}
	cout<<"Yes";
}