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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;

const int MAX_N=1e5;
int N,M;
typedef pair<int,int> P;
int v2g[MAX_N];
map<int,int> g2v[MAX_N];

int main(){
	cin>>N>>M;
	REP(i,N){
		v2g[i]=i;
		g2v[i].insert(P(i,0));
	}
	bool ans=true;
	REP(i,M){
		int L,R,D;
		cin>>L>>R>>D;
		L--;
		R--;
		if (v2g[L]==v2g[R]){
			int g=v2g[L];
			if (g2v[g][R]-g2v[g][L]!=D){
				ans=false;
			}
		}else{
			int f,t,d;
			if (g2v[v2g[L]].size()<g2v[v2g[R]].size()){
				f=v2g[L];
				t=v2g[R];
				d=-D-g2v[f][L]+g2v[t][R];
			}else{
				f=v2g[R];
				t=v2g[L];
				d=D-g2v[f][R]+g2v[t][L];
			}
			for(auto i:g2v[f]){
				g2v[t][i.first]=i.second+d;
				v2g[i.first]=t;
			}
		}
	}
	cout<<(ans ? "Yes" : "No")<<endl;
	return 0;
}