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>
#define pb push_back
#define mp make_pair
#define F first
#define S second

using namespace std;

typedef long long ll;
typedef pair<int , int> pii;

const int N = 100100;

vector <pii> adj[N];

bool mark[N];
int dist[N];

void DFS(int v){
	
	mark[v] = 1;

	for(auto u : adj[v]){
	   int t = dist[v] + u.S;
	   if(!mark[u.F]){
		  dist[u.F] = t;
		  DFS(u.F);
	   }
	   else if(dist[u.F] != t){
		  cout << "No" << endl;
		  exit(0);
	   }
	}
	
	return;
}

int main(){

	ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);

	int n , m , l , r , d;
	cin >> n >> m;

	for(int i = 0 ; i < m ; i++){
	   cin >> l >> r >> d;
	   adj[--l].pb({--r , d});
	   adj[r].pb({l , -d});
	}
	
	for(int i = 0 ; i < n ; i++)
	   if(!mark[i])
		  DFS(i);
	
	cout << "Yes" << endl;

	return 0;
}