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 <vector>
using namespace std;

vector<vector<int>> G;
int d[200010], l[200010], r[200010], x[200010];
bool det[200010];

bool dfs(int key){
	det[key]=true;
	bool res=true;
	for(int i=0; i<G[key].size(); ++i){
		int antic=x[key], nx=l[G[key][i]]+r[G[key][i]]-key;
		if(l[G[key][i]]==key) antic += d[G[key][i]];
		else antic -= d[G[key][i]];
		if(det[nx]){
			res=res&&(x[nx]==antic);
		}else{
			x[nx]=antic;
			res=res&&dfs(nx);
		}
	}
	return res;
}

int main() {
	int N, M;
	cin >> N >> M;
	G.resize(N+1);
	for(int i=0; i<M; ++i){
		cin >> l[i] >> r[i] >> d[i];
		G[l[i]].emplace_back(i);
		G[r[i]].emplace_back(i);
	}
	bool flag=true;
	for(int i=1; i<=N; ++i){
		if(!det[i]){
			x[i]=0;
			flag=flag&&dfs(i);
		}
	}
	cout << (flag ? "Yes" : "No") << endl;
	return 0;
}