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;
const int N=100005;
int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1; ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int f[N],d[N];
int find(int x){
	if (x==f[x]) return x;
	int fa=find(f[x]);
	d[x]=d[x]+d[f[x]];
	return f[x]=fa;
}
int main(){
	int n=read(),m=read();
	for (int i=1;i<=n;i++){
		d[i]=0; f[i]=i;
	}
	bool pd=0;
	while (m--){
		int l=read(),r=read(),v=read();
		int fl=find(l),fr=find(r);
		if (fl==fr){
			if (d[l]-d[r]!=v) pd=1;
		}else{
			f[fl]=fr;
			d[fl]=v+d[r]-d[l];
		}
	}
	if (pd==1) printf("No\n");
	else printf("Yes\n");
	return 0;
}