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: Memory Limit Exceeded)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
boolean ans=true;
		int n = sc.nextInt();
		int m = sc.nextInt();
		int l;
		int r;
		int d;
		int graph[][] = new int[n + 1][n + 1];
		for (int i = 0; i <= n; i++) {
			Arrays.fill(graph[i], 999999);
		}
		for (int i = 1; i <= m; i++) {
			l = sc.nextInt();
			r = sc.nextInt();
			d = sc.nextInt();
			graph[l][r] = d;
			graph[r][l] = (-1) * d;

		}
		ans = floydWarshall(graph);



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

	}


	static boolean floydWarshall(int graph[][]) {

		int V = graph[0].length;
boolean ans=true;
		int i, j, k;

		for (i = 0; i < V; i++) {
			for (j = 0; j < V; j++) {
				graph[i][j] = graph[i][j];
			}
		}
		for (k = 0; k < V; k++) {

			for (i = 0; i < V; i++) {

				for (j = 0; j < V; j++) {

					if (graph[i][k] + graph[k][j] < graph[i][j]) {
						if(graph[i][j]<=10000){
							ans=false;
						}
						graph[i][j] = graph[i][k] + graph[k][j];
					}
				}
			}
		}

		return ans;
	}

}