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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastScanner in = new FastScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		ProblemD solver = new ProblemD();
		solver.solve(1, in, out);
		out.close();
	}

	static class ProblemD {
		public void solve(int testNumber, FastScanner in, PrintWriter out) {
			int n = in.nextInt();
			List<Edge>[] edges = new List[n];
			for (int i = 0; i < n; i++) {
				edges[i] = new ArrayList<>();
			}
			int m = in.nextInt();
			for (int i = 0; i < m; i++) {
				int a = in.nextInt() - 1;
				int b = in.nextInt() - 1;
				int c = in.nextInt();
				edges[a].add(new Edge(b, c));
				edges[b].add(new Edge(a, -c));
			}
			int[] x = new int[n];
			boolean[] was = new boolean[n];
			for (int i = 0; i < n; i++) {
				if (!was[i]) {
					dfs(i, edges, x, was);
				}
			}
			for (int i = 0; i < n; i++) {
				for (Edge e : edges[i]) {
					if (x[e.dst] - x[i] != e.c) {
						out.println("No");
						return;
					}
				}
			}
			out.println("Yes");
		}

		private void dfs(int i, List<Edge>[] edges, int[] x, boolean[] was) {
			was[i] = true;
			for (Edge e : edges[i]) {
				int j = e.dst;
				if (was[j]) {
					continue;
				}
				x[j] = x[i] + e.c;
				dfs(j, edges, x, was);
			}
		}

		class Edge {
			int dst;
			int c;

			Edge(int dst, int c) {
				this.dst = dst;
				this.c = c;
			}

		}

	}

	static class FastScanner {
		private BufferedReader in;
		private StringTokenizer st;

		public FastScanner(InputStream stream) {
			in = new BufferedReader(new InputStreamReader(stream));
		}

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

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}
}