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:
/*input
3 3
1 2 1
2 3 1
1 3 5


*/
#include <iostream>
#include <vector>

typedef std::pair<int, int> ii;

const int ms = 100100;

std::vector<ii> edges[ms];
int pos[ms];
bool visit[ms];

bool dfs(int on, int cur = 0) {
	if(visit[on]) {
		return cur == pos[on];
	}
	pos[on] = cur;
	visit[on] = true;
	for(auto a : edges[on]) {
		if(!dfs(a.first, cur + a.second)) {
			return false;
		}
	}
	return true;
}

int main() {
	int n;
	int m;
	std::cin >> n >> m;
	for(int i = 0; i < m; i++) {
		int u, v, d;
		std::cin >> u >> v >> d;
		edges[u].push_back(ii(v, d));
		edges[v].push_back(ii(u, -d));
	}
	bool ans = true;
	for(int i = 1; i <= n; i++) {
		if(!visit[i]) {
			ans = ans && dfs(i);
		}
	}
	std::cout << (ans ? "Yes" : "No") << '\n';
}