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.*;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        int n = in.nextInt(), m = in.nextInt();
        int[] pos = new int[n];

        Arrays.fill(pos, Integer.MAX_VALUE);

        HashMap<Integer, List<int[]>> hm = new HashMap<>();
        for (int i = 0 ; i < m; i++) {
            int l = in.nextInt() - 1, r = in.nextInt() - 1, d = in.nextInt();
            List<int[]> list = hm.getOrDefault(l, new ArrayList<>());
            list.add(new int[]{l, r, d});
            hm.put(l, list);
            List<int[]> list2 = hm.getOrDefault(r, new ArrayList<>());
            list2.add(new int[]{l, r, d});
            hm.put(r, list2);
            if (i == 0) {
                pos[l] = 0;
            }
        }
        boolean set = true;
        while (hm.size() > 0) {
            boolean update = false;
            
            for (int j = 0; j < n; j++) {
//                System.out.println(j);
                if (hm.containsKey(j) && set) {
                    pos[j] = 0;
                    set = false;
                }

                if (!hm.containsKey(j) || pos[j] == Integer.MAX_VALUE)
                    continue;
                List<int[]> todo = hm.get(j);
                for (int i = 0; i < todo.size(); i++) {
                    int[] aaa = todo.get(i);
                    int l = aaa[0], r = aaa[1], d = aaa[2];
                    if (pos[l] == Integer.MAX_VALUE) {
                        pos[l] = pos[r] - d;
                    } else if (pos[r] == Integer.MAX_VALUE) {
                        pos[r] = pos[l] + d;
                    } else {
                        if (pos[r] != pos[l] + d) {
                            System.out.println("No");
                            return;
                        }
                    }
                }
                update = true;
                hm.remove(j);
            }
            if (hm.size() != 0 && !update) {
                set = true;
            }
        }
        System.out.println("Yes");
    }
}