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 JAVA):

Solution:
import java.util.*;

public class Main {

	static int N;
	static HashMap<Integer, HashMap<Integer, Integer>> edges;
	static int[] values;
	static boolean[] visited;
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);

		N = in.nextInt();
		int M = in.nextInt();
		
		edges = new HashMap<Integer, HashMap<Integer, Integer>>();
		Input[] inputs = new Input[M];
		
		for (int i = 0; i < M; i++) {
			int l = in.nextInt();
			int r = in.nextInt();
			int c = in.nextInt();
			
			if (edges.containsKey(l)) {
				HashMap<Integer, Integer> e = edges.get(l);
				e.put(r, c);
			} else {
				HashMap<Integer, Integer> e = new HashMap<Integer, Integer>();
				e.put(r, c);
				edges.put(l, e);
			}
			
			if (edges.containsKey(r)) {
				HashMap<Integer, Integer> e = edges.get(r);
				e.put(l, -1 * c);
			} else {
				HashMap<Integer, Integer> e = new HashMap<Integer, Integer>();
				e.put(l, -1 * c);
				edges.put(r, e);
			}
			
			inputs[i] = new Input(l, r, c);
		}
		
		values = new int[N + 1];
		visited = new boolean[N + 1];
		
		for (int i = 1; i < N + 1; i++) {
			if (visited[i] == false) {
				dfs(i, 0);
			}
		}
		
		boolean answer = true;
		
		for (int i = 0; i < M && answer; i++) {
			if (values[inputs[i].l] + inputs[i].c != values[inputs[i].r]) {
				answer = false;
			}
		}
		
		System.out.println(answer ? "Yes" : "No");
	}
	
	public static void dfs(int n, int val) {
		visited[n] = true;
		values[n] = val;
		
		HashMap<Integer, Integer> e = edges.get(n);
		
		if (e == null) {
			return;
		}
		
		for (int next : e.keySet()) {
			if (visited[next] == false) {
				dfs(next, val + e.get(next));
			}
		}
	}
}

class Input {
	int l;
	int r;
	int c;
	
	public Input(int l, int r, int c) {
		this.l = l;
		this.r = r;
		this.c = c;
	}
}