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 main() {
	ios::sync_with_stdio(0), cin.tie(0);
	int n, m;
	cin >> n >> m;
	vector<vector<pair<int,int>>> edges(n);
	for(int i = 0; i < m; i++){
		int l, r, d;
		cin >> l >> r >> d;
		l--; r--;
		edges[l].emplace_back(r, d);
		edges[r].emplace_back(l, -d);
	}
	vector<int> done(n);
	vector<int> x(n);
	for(int i = 0; i < n; i++){
		if(done[i]) continue;
		done[i] = 1;
		vector<int> bfs{i};
		for(int s = 0; s < int(bfs.size()); s++){
			int v = bfs[s];
			for(auto z : edges[v]){
				int w = z.first;
				int nx = z.second + x[v];
				if(!done[w]){
					done[w] = 1;
					x[w] = nx;
					bfs.push_back(w);
				} else if(x[w] != nx){
					cout << "No" << '\n';
					exit(0);
				}
			}
		}
	}
	cout << "Yes" << '\n';
}