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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    static int[] dis,pre;
    
    static int  find(int x) {
    	while(x!=pre[x]) {
    		int r = pre[x];
//    		dis[x] += dis[r];
    		x = r;
    	}
    	return x;
    }
    
    static void Merge(int x,int y,int d) {
    	int fx = find(x);
    	int fy = find(y);
    	if(fx != fy) {
    		pre[fy] = pre[fx];
    		dis[y] += d + dis[x];
    	}
    }
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		
		dis = new int[100005];
		pre = new int[100005];
		
		int n = scanner.nextInt();
		int m = scanner.nextInt(); 
		boolean ans = true;
		for(int i=1;i<=n;i++) pre[i] = i;
		for(int i=0;i<m;i++) {
			int x = scanner.nextInt();
			int y = scanner.nextInt();
			int d = scanner.nextInt();
//			System.out.println("x="+x+"  y="+y);
//			System.out.println("pre[x]="+pre[x]+"  pre[y]="+pre[y]);
			if(pre[x]!=pre[y]) {
				Merge(x,y,d);
			}else {
				if(dis[y]!=d) {ans = false; break;}
			}
		}
//		for(int i=1;i<=n;i++) System.out.println("dis["+i+"]="+dis[i]);
		if(ans) System.out.println("Yes");
		else System.out.println("No");
		
	}

}