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<iostream>
#include<vector>
#include<stack>
using namespace std;

int main() {
	int n, m; cin >> n >> m;
	vector<int>x(n, 0); stack<int>s;
	vector < vector<pair<int, int>>>l(n), r(n);
	for (int i = 0, L, R, d; i < m; i++) {
		cin >> L >> R >> d; L--; R--;
		l[R].push_back({ L,d }); r[L].push_back({ R,d });
	}
	for (int i = 0; i < n; i++) {
		if (l[i].size() == 0) { x[i] = 1; s.push(i); }
	}
	bool ok = true;
	while (!s.empty()) {
		int node = s.top(); s.pop();
		for (int i = 0; i < r[node].size(); i++) {
			int next = r[node][i].first, xd = x[node] + r[node][i].second;
			if (x[next] == 0) { x[next] = xd; s.push(next); }
			else if (x[next] != xd) { ok = false; break; }
		}
	}
	for (int i = 0; i < n; i++) if (x[i] == 0)ok = false;
	cout << (ok ? "Yes" : "No") << endl;
	return 0;
}