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>
using namespace std;
#define ll int64_t

typedef pair<int,int> P;

const int INF=numeric_limits<int>::max();
const int N=1e5+5;
int n,m,dist[N];
vector<P> adj[N];

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m;
	for(int i=0;i<m;++i){
		int l,r,d;
		cin>>l>>r>>d;
		adj[l].emplace_back(r,d);
		adj[r].emplace_back(l,-d);
	}
	for(int i=1;i<=n;++i) dist[i]=INF;
	for(int i=1;i<=n;++i)
		if(dist[i]==INF){
			dist[i]=0;
			queue<int> Q;
			Q.push(i);
			while(!Q.empty()){
				int u=Q.front();
				Q.pop();
				for(P nxt:adj[u]){
					int v=nxt.first;
					int d=nxt.second;
					if(dist[v]==INF){
						dist[v]=dist[u]+d;
						Q.push(v);
					}
					else if(dist[v]!=dist[u]+d){
						return cout<<"No\n",0;
					}
				}
			}
		}
	cout<<"Yes\n";
}