Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

class Edge {
	Integer node;
	Integer d;

	Edge(Integer _n, Integer _d) {
		node = _n;
		d = _d;
	}
}

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	boolean dfs(int idx, int newPos) {

		// 訪問済みノードの場合、記録済みの座標と新しい座標のチェックをする。（異なっていれば矛盾あり）
		if (!notVisited.contains(idx)) {
			if (pos.get(idx) == newPos) {
				return true;
			} else {
				return false;
			}
		}

		// 座標を記録し、未訪問ノードから削除する
		pos.put(idx, newPos);
		notVisited.remove(idx);

		int currentPos = newPos;
		List<Edge> edges = g.get(idx);
		for (Edge e : edges) {
			if (!dfs(e.node, currentPos + e.d)) {
				return false;
			}
		}
		return true;
	}

	Map<Integer, List<Edge>> g = new HashMap<>();
	Set<Integer> notVisited;
	Map<Integer, Integer> pos = new HashMap<>();

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();
		int M = sc.nextInt();

		for (int i = 0; i < M; i++) {
			final int l = sc.nextInt();
			final int r = sc.nextInt();
			final int d = sc.nextInt();

			g.compute(l, (k, v) -> {
				if (v == null) {
					List<Edge> list = new ArrayList<>();
					list.add(new Edge(r, d));
					return list;
				} else {
					v.add(new Edge(r, d));
					return v;
				}
			});
			g.compute(r, (k, v) -> {
				if (v == null) {
					List<Edge> list = new ArrayList<>();
					list.add(new Edge(l, -d));
					return list;
				} else {
					v.add(new Edge(l, -d));
					return v;
				}
			});

		}

		sc.close();

		notVisited = new HashSet<>(g.keySet()); // 複製

		String ans = "Yes";

		Iterator<Integer> it;
		while ((it = notVisited.iterator()).hasNext()) {
			int idx = it.next();
			if (!dfs(idx, 0)) {
				ans = "No";
				break;
			}
		}

		out.println(ans);

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}

}