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;
vector<vector<pair<int,int>>> A;
vector<bool> visited;
vector<int> pos;
bool g(int p, int q) {
	if (visited[p]) {
		return pos[p] == q;
	} else {
		visited[p] = true;
		pos[p] = q;
		for (auto x : A[p]) {
			if (!g(x.first, q + x.second)) return false;
		}
		return true;
	}
}
bool f() {
	visited.resize(N);
	pos.resize(N);
	for (int i = 0; i < N; ++ i) if (!visited[i]) {
		if (!g(i, 0)) return false;
	}
	return true;
}
int main() {
	cin >> N >> M;
	A.resize(N);
	for (int i = 0; i < M; ++ i) {
		int Li, Ri, Di;
		cin >> Li >> Ri >> Di;
		-- Li; -- Ri;
		A[Li].push_back({Ri, Di});
		A[Ri].push_back({Li, -Di});
	}

	cout << (f() ? "Yes" : "No") << endl;
}