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 size=100010;
int n,m;
int used[size];
int d[size];
bool flag=false;
struct edge//from left to right
{
 	int to,val;
 	edge(int to=0,int val=0):to(to),val(val) {}
 };

vector<edge> V[size];

void dfs(int from,int to,int len)
{
 	if(used[to])
 		if(d[from]+len!=d[to])
 		{
		 	cout<<"No"<<endl;
		 	exit(0);
			}
		else return;
		
	used[to]=1;
	d[to]=d[from]+len;
	for(int i=0;i<V[to].size();i++)
	{
		int nto=V[to][i].to;
		dfs(to,nto,V[to][i].val);
		}
	} 

int main()
{
	cin>>n>>m;
	if(m==0)
	{
		cout<<"Yes"<<endl;
		return 0;
		}	
	
	for(int i=0;i<m;i++)
	{
		int L,R,len;
		cin>>L>>R>>len;
		V[L].push_back(edge(R,len));
		V[R].push_back(edge(L,-len));
		}
	for(int i=1;i<=n;i++)
		if(!used[i])
			dfs(i,i,0);
	cout<<"Yes"<<endl;		
 	return 0;
 }