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.util.*;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        ArrayList<LinkedList<Tuple>> list = new ArrayList<>();
        for(int i = 0; i <= n; i++) list.add(new LinkedList<>());
        int[] memo = new int[3];
        for (int i = 0; i < m; i++) {
            memo[0] = sc.nextInt();
            memo[1] = sc.nextInt();
            memo[2] = sc.nextInt();
            list.get(memo[0]).add(new Tuple(memo[1], memo[2]));
            list.get(memo[1]).add(new Tuple(memo[0], -memo[2]));
        }
        // DFS
        Deque<Integer> deque = new ArrayDeque<>();
        boolean[] visit = new boolean[n + 1];
        int[] distance = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            if (!visit[i]) {
                distance[i] = 0;
                visit[i] = true;
                deque.push(i);
                while (!deque.isEmpty()) {
                    int curr = deque.poll();
                    Iterator<Tuple> it = list.get(curr).iterator();
                    while(it.hasNext()) {
                        Tuple next = it.next();
                        if (!visit[next.x]) {
                            visit[next.x] = true;
                            distance[next.x] = distance[curr] - next.y;
                            deque.push(next.x);
                        } else {
                            if (distance[next.x] != distance[curr] - next.y) {
                                System.out.println("No");
                                return;
                            }
                        }
                    }
                }
            }
        }
        System.out.println("Yes");
    }

    static class Tuple {
        final int x;
        final int y;
        Tuple(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
}