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.util.*;

public final class Main {


    static int N;

    static int M;



    static class Node {

        List<Edge> childs = new LinkedList<>();

        int index;

        public void addChild(Node n , int num) {

            Edge e=  new Edge(n,num);

            childs.add(e);
        }

        public Node(int index) {

            this.index = index;
        }

    }

    static class Edge {

        Node from;

        Node to;

        int num;

        public Edge( Node to ,int num) {

            this.to = to;

            this.num = num;
        }
    }

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);


        N = s.nextInt();

        M = s.nextInt();

        Node[] nodes = new Node[N+1];


        for (int i = 0; i  < M; i++) {

            int a = s.nextInt();

            int b = s.nextInt();

            int w = s.nextInt();

            if (nodes[a] != null && nodes[b] != null) {

                int result1 = dfs(nodes[a], b);

                if (result1 != -1) {

                    if (result1 != w) {

                        System.out.print("No");
                        return;
                    }

                    nodes[a].addChild(nodes[b] , w);
                }

                int result2 = dfs(nodes[b], a);

                if (result2 != -1) {

                    System.out.print("No");

                    return;
                }

            } else {

                if (nodes[a] == null) nodes[a] = new Node(a);

                if (nodes[b] == null) nodes[b] = new Node(b);

                nodes[a].addChild(nodes[b],w);
            }




        }

        System.out.print("Yes");

    }


    public  static int dfs(Node n1, int index) {

        if (n1.index == index ) return 0;

        else if (n1.childs.isEmpty()) return -1;

        int result = -1;

        for (Edge e : n1.childs) {

            Node k = e.to;

            int localw = e.num;

            int re=dfs(k, index);

            if (re != -1)  result = re+ e.num;
        }

        return result;
    }

}