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 class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        WeightedUnionFind wuf = new WeightedUnionFind(n);
        for(int i = 0; i < m; i++){
            if(!wuf.union(sc.nextInt()-1,sc.nextInt()-1,sc.nextLong())){
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
    }
}

class WeightedUnionFind{
    int[] parent;
    int[] rank;
    long[] weight;
    int[] size;

    public WeightedUnionFind(int size) {
		this.parent = new int[size];
		this.rank = new int[size];
        this.weight = new long[size];
		this.size = new int[size];
		
		for (int i = 0; i < size; i++) {
			makeSet(i);
		}
	}
	
	public void makeSet(int i) {
		parent[i] = i;
		rank[i] = 0;
		size[i] = 1;
	}

    public boolean union(int x, int y, long w) {
        w += getWeight(x);
        w -= getWeight(y);
		int xRoot = find(x);
		int yRoot = find(y);
        if(xRoot == yRoot){
            return getDistance(xRoot,yRoot) == w;
        } else if (rank[xRoot] > rank[yRoot]) { 
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
            weight[yRoot] = w;
		} else if(rank[xRoot] < rank[yRoot]) {
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
            weight[xRoot] = -w;
		} else if (xRoot != yRoot){
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
			rank[xRoot]++; 
            weight[yRoot] = w;
		}
        return true;
	}
	
	public int find(int i) {
		if (i != parent[i]) {
		    parent[i] = find(parent[i]);
            weight[i] += weight[parent[i]];
		}
		return parent[i];
	}

    public long getWeight(int i){
        find(i);
        return weight[i];
    }

     public long getDistance(int x, int y) {
        return getWeight(y) - getWeight(x);
    }
	
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}

	public int size(int x) {
		return size[find(x)];
	}
}