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<iostream>
#include<cstdio>
#include<vector>
using namespace std;

int p[5],v[100005],o[100005],f=1;
vector<int>b[100005];

void dfs(int a,int d){
	int i;
	//cout<<a<<' '<<d<<endl;
	if(o[a]!=0){
		if(v[a]!=d)f=0;
		return;
	}
	
	v[a]=d;
	o[a]=1;
	
	for(i=0;i<b[a].size();i+=2){
		dfs(b[a][i],d+b[a][i+1]);
	}
	
	return;
}

int main(){
	int n,m,i;
	cin>>n>>m;
	
	for(i=1;i<=m;i++){
		cin>>p[1]>>p[2]>>p[3];
		b[p[1]].push_back(p[2]);
		b[p[1]].push_back(p[3]);
		b[p[2]].push_back(p[1]);
		b[p[2]].push_back(0-p[3]);
	}
	
	for(i=1;i<=n;i++){
		if(o[i]==0)dfs(i,0);
	}
	
	if(f==1)
		cout<<"Yes"<<endl;
	else
		cout<<"No"<<endl;
	
	return 0;
}