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<cstring>

namespace fuck
{
	const int N=101000,M=N*4;
	int begin[N],next[M],to[M],w[M];
	int n,m,e;
	void add(int x,int y,int z,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		w[e]=z;
		if(k)add(y,x,-z,0);
	}
	void initialize()
	{
		scanf("%d%d",&n,&m);
		for(int i=1,u,v,k;i<=m;i++)
		{
			scanf("%d%d%d",&u,&v,&k);
			add(u,v,k);
		}
	}

	bool vis[N];
	int f[N];

	bool dfs(int p,int x)
	{
		if(vis[p])return f[p]==x;

		vis[p]=1,f[p]=x;

		for(int i=begin[p];i;i=next[i])
			if(!dfs(to[i],x+w[i]))return 0;
		return 1;
	}

	bool judge()
	{
		for(int i=1;i<=n;i++)
			if(!vis[i] && !dfs(i,0))
				return 0;
		return 1;
	}

	void solve()
	{
		initialize();
		printf("%s\n",judge()?"Yes":"No");
	}
}

int main()
{
	fuck::solve();
	return 0;
}