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 n,m,dis[200001];
bool vis[200001];
vector<pair<int,int> > g[200001];

bool dfs(int x){
	int i,y,d;
	vis[x]=true;
	for(i=0;i<g[x].size();i++){
		y=g[x][i].first,d=g[x][i].second;
		if(!vis[y]){
			dis[y]=dis[x]+d;
			if(!dfs(y))return false;
		}	
		else
			if(dis[x]+d!=dis[y])return false;
	}
	return true;
}

int main(){
	int i,u,v,w;
	cin >> n >> m;
	for(i=1;i<=m;i++){
		cin >> u >> v >> w;
		g[u].push_back(make_pair(v,w));
		g[v].push_back(make_pair(u,-w));
	}
	for(i=1;i<=n;i++)
		if(!vis[i])
			if(!dfs(i)){
				cout << "No" << endl;
				return 0;
			}
	cout << "Yes";
	return 0;
}