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 ) {
		Scanner sc			= new Scanner( System.in ) ;
		int		N			= sc.nextInt();
		int		M			= sc.nextInt();
		int		field[][]	= new int[M][3];

		for( int q = 0; q < M; q++ ) {
			for( int p = 0; p < 3; p++ ) {
				field[q][p] = sc.nextInt() ;
			}
		}

		int		matrix[][]		= new int[N][N] ;
		for( int q = 0; q < N; q++ ) {
			for( int p = 0; p < N; p++ ) {
				matrix[q][p] = 0 ;
			}
		}
		
		for(int data = 0; data < M; data++) {
			matrix[field[data][0]-1][field[data][1]-1] = - field[data][2] ;
			matrix[field[data][1]-1][field[data][0]-1] =   field[data][2] ;
		}
		
		int max = 0 ;
		for( int q = 0; q < N; q++ ) {
			for( int p = 0; p < N; p++ ) {
				for( int index = 0; index < N; index++ ) {
					//System.out.println( matrix[q][index] + " + " + matrix[index][p] + " = " + matrix[q][p] ) ;
					if( matrix[q][index] + matrix[index][p] != matrix[q][p] ) {
						System.out.println("No") ;
						return ;
					}
					if( matrix[q][p] > max ) {
						max = matrix[q][p] ;
					}
				}
			}
		}
		
		if( max > 1_000_000_000 ) {
			System.out.println("No") ;
			return ;
		}
		
		System.out.println("Yes") ;
	}
}