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)

/*
 * x 軸上に N 人の人が立っています。 人 i の位置を xi とします。 任意の i に対して、xi は 0 以上 109 以下の整数です。 同じ位置に複数の人が立っていることもありえます。
 * 
 * これらの人の位置に関する情報が M 個与えられます。 このうち i 個めの情報は (Li,Ri,Di) という形をしています。 
 * この情報は、人 Ri は人 Li よりも距離 Di だけ右にいること、 すなわち、xRi−xLi=Di が成り立つことを表します。
 * 
 * これら M 個の情報のうちのいくつかに誤りがある可能性があることがわかりました。 与えられる M 個すべての情報と矛盾しないような値の組 (x1,x2,…,xN) が存在するかどうか判定してください。
 */

import java.util.*;
	public class Main{
        public static void main(String[] args){
            Scanner sc = new Scanner(System.in);
            //入力
            int n = Integer.parseInt(sc.next());
            int m = Integer.parseInt(sc.next());
            int[] states = new int[n];
            int[] flags = new int[n];
            int[][] distance = new int[m][3];
            int flag = 0;
            int[] flag_data = new int[m];
            int flagDataCounter = 0;
            int updateCounter = 0;
            
            for(int i=0; i<m; i++){
            	distance[i][0] = Integer.parseInt(sc.next())-1;
            	distance[i][1] = Integer.parseInt(sc.next())-1;
            	distance[i][2] = Integer.parseInt(sc.next());
            	flag_data[i] = 0;
            }
            for(int i=0; i<n; i++){
            	flags[i] = 0;
            }

            while(flag == 0 && flagDataCounter < m){
            	if(updateCounter == 0){
            		for(int i=1; i<m; i++){
                		if(flag_data[i] == 0){
                			flags[distance[i][0]] = 1;
                			states[distance[i][0]] = 0;
                			break;
                		}
                	}
            		updateCounter = 1;
            	}else{
                	for(int i=0; i<m; i++){
                		updateCounter = 0;
                		if(flags[distance[i][0]] == 1 && flags[distance[i][1]] == 0){
                			states[distance[i][1]] = states[distance[i][0]] + distance[i][2];
                			flags[distance[i][1]] = 1;
                			updateCounter++;
                			flag_data[i] = 1;
                			flagDataCounter++;
                		}else if(flags[distance[i][0]] == 0 && flags[distance[i][1]] == 1){
                			states[distance[i][0]] = states[distance[i][1]] - distance[i][2];
                			flags[distance[i][0]] = 1;
                			updateCounter++;
                			flag_data[i] = 1;
                			flagDataCounter++;
                		}else if(flags[distance[i][0]] * flags[distance[i][1]] == 1 && flag_data[i] == 0){
                			if(states[distance[i][1]] - states[distance[i][0]] != distance[i][2]){
                				flag = 1;
                				break;
                			}else{
                				flag_data[i] = 1;
                				flagDataCounter++;
                			}
                		}
                	}
            	}
            }
            
            //出力
            if(flag == 1){
            	System.out.println("No");
            }else{
            	System.out.println("Yes");
            }
            sc.close();
        }
    }