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;

int n, m, l, r, d1;
vector<pair<int,int>> e[100004];
bool visited[100004];
long long d[100004];
bool ok;
void dfs(int here){
	visited[here] = true;
	for(auto i : e[here]){
		int there = i.first;
		int dist = i.second;
		if(!visited[there]){	
			d[there] = d[here] + dist;
			dfs(there);
		}
		else
			if(d[here]+dist != d[there])	
				ok = false;
		}
	return;
}

int main() {
	cin>>n>>m;
	while(m--){
		cin>>l>>r>>d1;
		e[l].push_back({r,d1});
		e[r].push_back({l,-d1});
	}
	ok = true;
	memset(visited,false,sizeof(visited));
	memset(d,0,sizeof(d));
	for(int i = 1; i<=n;i++){
		if(visited[i])	continue;
		d[i] = 0;
		dfs(i);
	}
	cout<<(ok? "Yes" : "No");
	return 0;
}