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);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] position = new int[N];
		for (int i=0;i<N;i++) {
			position[i] = -1;
		}
		for (int i=0;i<M;i++) {
			int positionA ,positionB,D;
			positionA = sc.nextInt();
			positionB = sc.nextInt();
			D = sc.nextInt();
			if(position[positionA-1]==-1) {
				if (position[positionB-1]==-1) {
					position[positionA-1] = 500000000;
					position[positionB-1] = D+500000000;
				} else {
					int tempA = position[positionB-1]-D;
					if (tempA<0) {
						for (int j=0;j<N;j++) {
							if (position[j]!=-1) {
								position[j]+=Math.abs(tempA);
								if (position[j]>1000000000) {
									System.out.println("No");
									return ;
								}
							}
						}
						position[positionA-1] = 0;
					} else {
						position[positionA-1] = tempA;
					}
				}
			} else if (position[positionB-1]==-1) {
				int tempB = position[positionA-1]+D;
				if (tempB>1000000000) {
					for (int j=0;j<N;j++) {
						if (position[j]!=-1) {
							position[j]-=tempB -1000000000;
							if (position[j]<0) {
								System.out.println("No");
								return ;
							}
						}
					}
					position[positionB-1] = 1000000000;
				} else {
					position[positionB-1] = tempB;
				}
			} else {
				if (position[positionB-1]-position[positionA-1]!=D) {
					System.out.println("No");
					return ;
				}
			}
		}
		System.out.println("Yes");
	}
}