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;
const int N = 1e5 + 10, M = 2e5 + 10, inf = 1e9 + 1;
int dis[N], mark[N], n, m, mn, mx;
vector<pair<int, int> > gr[N];
void dfs(int v) {
	mark[v] = 1;
	for (auto p : gr[v]) {
		int u = p.first, w = p.second;
		if (mark[u]) {
			if (dis[u] != dis[v] + w) {
				cout << "No";
				exit(0);
			}
		}
		else {
			dis[u] = dis[v] + w;
			dfs(u);
		}
	}
	return;
}
int main() {
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int x, y, w;
		cin >> x >> y >> w;
		gr[x].push_back({y, w});
		gr[y].push_back({x, -w});
	}
	for (int i = 0; i < n; i++) {
		if (!mark[i]) {
			mn = mx = 0;
			dfs(i);
			if (mx - mx > inf) {
				cout << "No";
				return 0;
			}
		}
	}
	cout << "Yes";
	return 0;
}