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:
#prag\
ma GCC optimize("Ofast")
#include <bits/stdc++.h>
typedef long long ll;
#define N 200005
using namespace std;
int n,m,dis[N];
bool visited[N];
vector <pair<int,int > > V[N]; 
bool dfs(int x)
{
	visited[x]=true;
	for(int i=0;i<V[x].size();++i)
	{
		int y=V[x][i].first,d=V[x][i].second;
		if(!visited[y])
		{
			dis[y]=dis[x]+d;
			if(!dfs(y)) return false;
		}	
		else if(dis[x]+d!=dis[y]) return false;
	}
	return true;
} 
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;++i)
	{
		int u,v,w; scanf("%d%d%d",&u,&v,&w);
		V[u].push_back(make_pair(v,w));
		V[v].push_back(make_pair(u,-w));
	}
	for(int i=1;i<=n;++i)
		if(!visited[i])
			if(!dfs(i))
			{
				printf("No\n");
				return 0;
			}
	printf("Yes\n");
	return 0;
}