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.io.*;
import java.util.*;

public class Main {

	ArrayList<Integer>[] g;
	
	boolean[] vis;
	int[] delta;
	
	void submit() {
		int n = nextInt();
		int m = nextInt();
		
		g = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
		}
		
		while (m-- > 0) {
			int v = nextInt() - 1;
			int u = nextInt() - 1;
			int d = nextInt();
			g[v].add(u);
			g[v].add(d);
			g[u].add(v);
			g[u].add(-d);
		}
		
		vis = new boolean[n];
		delta = new int[n];
		
		for (int i = 0; i < n; i++) {
			if (!vis[i]) {
				if (!dfs(i, 0)) {
					out.println("No");
					return;
				}
			}
		}
		
		out.println("Yes");
	}
	
	boolean dfs(int v, int x) {
		vis[v] = true;
		delta[v] = x;
		
		for (int i = 0; i < g[v].size(); i += 2) {
			int u = g[v].get(i);
			int d = g[v].get(i + 1);
			if (!vis[u]) {
				if (!dfs(u, x + d)) {
					return false;
				}
			} else {
				if (delta[u] != x + d) {
					return false;
				}
			}
		}
		
		return true;
	}

	void preCalc() {

	}

	void stress() {

	}

	void test() {

	}

	Main() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		preCalc();
		submit();
		//stress();
		//test();
		out.close();
	}

	static final Random rng = new Random();

	static int rand(int l, int r) {
		return l + rng.nextInt(r - l + 1);
	}

	public static void main(String[] args) throws IOException {
		new Main();
	}

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			throw new RuntimeException(e);
		}
	}

	int nextInt() {
		return Integer.parseInt(nextToken());
	}

	long nextLong() {
		return Long.parseLong(nextToken());
	}

	double nextDouble() {
		return Double.parseDouble(nextToken());
	}
}