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: Wrong Answer)

import java.io.IOException;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class Main {

	static int n, m;

	public static void main(String[] args) throws IOException {

		Scanner in = new Scanner(System.in);
		List<List<Vertex>> G = new ArrayList<>(100010);
		n = in.nextInt();
		m = in.nextInt();
		for (int i = 0; i <= n; i++) {
			List<Vertex> v = new ArrayList<>();
			G.add(v);
		}

		// solve
		for (int i = 0; i < m; i++) {
			int l = in.nextInt();
			int r = in.nextInt();
			int d = in.nextInt();

			// l から r へのコストを取得する。経路が存在しない場合はそのまま追加する。
			// dfsないしはbfsで探索する
			int cost = bfs(G, l, r);

			// 取得したコストと d が一致する場合のみ追加
			if (cost == -1 || cost == d) {
				G.get(l).add(new Vertex(r, d));
				G.get(r).add(new Vertex(l, (-1)*d));
			} else {
				System.out.println("No");
				return;
			}
		}
		System.out.println("Yes");

	}
	static int bfs(List<List<Vertex>> G, int fromNode, int targetNodeNumber) {
		int cost = 0;
		boolean[] visit = new boolean[n+1];
		Queue<List<Vertex>> q = new ArrayDeque<>();
		q.add(G.get(fromNode));
		visit[fromNode] = true;

		while (!q.isEmpty()) {
			// 隣接ノードに立ち寄る
			List<Vertex> toList = q.remove();
			for (Vertex vertex : toList) {
				if (!visit[vertex.nodeNumber] && !visit[targetNodeNumber]) {
					q.add(G.get(vertex.nodeNumber));
					visit[vertex.nodeNumber] = true;
					cost += vertex.weight;
				}
			}
		}
		cost = visit[targetNodeNumber] ? cost : -1;
		return cost;
	}

	static class Vertex {
		int nodeNumber;
		int weight;

		public Vertex(int nodeNumber, int weight) {
			this.nodeNumber = nodeNumber;
			this.weight = weight;
		}
	}

}