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;
//#define int long long 
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define F first
#define S second
int const M=2e5+10,inf=1e9+10,mod=1e9+7,M2=1e6+10,M3=3e3+10;
int l[M],r[M],q[M],d[M];
vector<int>adj[M],edg[M],adj2[M],edg2[M];
bool mark[M];
int cur[M];
void dfs(int v)
{
	mark[v]=1;
	for(int i=0;i<adj[v].size();i++)
	{
		int u=adj[v][i];
		if(mark[u] && cur[u]!=edg[v][i]+cur[v])
		{
			cout<<"No";
			exit(0);
		}
		if(!mark[u])cur[u]=cur[v]+edg[v][i],dfs(u);
	}
}
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=m;i++)scanf("%d%d%d",&l[i],&r[i],&d[i]),adj[l[i]].pb(r[i]),adj[r[i]].pb(l[i]),edg[l[i]].pb(d[i]),edg[r[i]].pb(-d[i]);	
	int tail=0,head=0;
	int cnt=0;
	for(int i=1;i<=n;i++)
	{
		if(!mark[i])dfs(i);
	}
	cout<<"Yes";
}