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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int n = sc.nextInt();
		int m = sc.nextInt();
		Union union = new Union(n);
		int x, y, d;
		for (int i = 0; i < m; i++) {
			x = sc.nextInt() - 1;
			y = sc.nextInt() - 1;
			d = sc.nextInt();
			if (union.unite(x, y, d)) {

			} else {
				System.out.println("No");
				return;
			}
		}
		System.out.println("Yes");
	}
}

class Union {
	private int[] parent;
	private int[] cost;

	public Union(int n) {
		parent = new int[n];
		cost = new int[n];
		for (int i = 0; i < n; i++) {
			parent[i] = i;
			cost[i] = Integer.MAX_VALUE;
		}
	}

	public int root(int n) {
		if (parent[n] == n) {
			return n;
		}
		return parent[n] = root(parent[n]);
	}

	public boolean unite(int a, int b, int d) {
		int aroot = root(a);
		int broot = root(b);
		if (aroot == broot) {
			if (cost[b] == Integer.MAX_VALUE || cost[a] == Integer.MAX_VALUE) {
				if (cost[b] == d) {
					return true;
				} else {
					return false;
				}
			} else if (cost[b] - cost[a] == d) {
				return true;
			} else {
				return false;
			}
		}
		parent[broot] = aroot;
		if (cost[a] == Integer.MAX_VALUE) {
			cost[broot] = d;
		} else {
			cost[broot] = cost[a] + d;
		}
		return true;
	}

}