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:
#prag\
ma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define f first
#define s second
pair<int,int> fa[100001];

int find(int x){
	if(fa[x].f==x&&fa[x].s==0) return x;
	int t=find(fa[x].f);
	fa[x]=make_pair(t,fa[fa[x].f].s+fa[x].s);
	fa[x].f=t;
	return t;
}

int main()
{
	ios::sync_with_stdio(false);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;++i) fa[i]=make_pair(i,0);
	for(int i=1;i<=m;++i){
		int l,r,d;
		cin>>l>>r>>d;
		int x=find(l),y=find(r);
		if(x!=y){
			if(x>y) fa[x]=make_pair(y,-(fa[l].s-fa[r].s+d));
			else fa[y]=make_pair(x,fa[l].s-fa[r].s+d);
		}
		else{
			if(fa[r].s-fa[l].s!=d){
				cout<<"No";
				return 0;
			}
		}
	}
	cout<<"Yes";
}