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[]) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        Map<Integer, Integer>[] graph = new HashMap[n + 1];

        for(int i = 1; i <= n; i++)
            graph[i] = new HashMap<>();

        for(int i = 0; i < m; i++) {
            int from = sc.nextInt();
            int to = sc.nextInt();
            int w = sc.nextInt();

            graph[from].put(to, w);
            graph[to].put(from, -w);
        }

        boolean[] visited = new boolean[n + 1];
        long[] weights = new long[n + 1];

        for(int i = 1; i <= n; i++) {
            if(visited[i]) continue;
            if(!dfs(graph, visited, weights, i, 0)) {
                System.out.println("No");
                return;
            }
        }
        
        System.out.println("Yes");
    }

    private static boolean dfs(Map<Integer, Integer>[] graph, boolean[] visited, long[] weights, int node, int prevWeight) {
        if(visited[node]) {
            return weights[node] == prevWeight;
        }

        weights[node] = prevWeight;
        visited[node] = true;

        for(Map.Entry<Integer, Integer> entry : graph[node].entrySet()) {
            int to = entry.getKey();
            int w = entry.getValue();

            if(!dfs(graph, visited, weights, to, w + prevWeight))
                return false;
        }
        return true;
    }
}