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 {
	static int[] states;
	static int[] belongs;
	public static void main(String[] args) {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		states = new int[n];
		belongs = new int[n];
		for (int i = 0; i < n; i++) {
			belongs[i] = i;
			states[i] = 0;
		}
		for(int i = 0; i < m; i++) {
			int l = sc.nextInt()-1;
			int r = sc.nextInt()-1;
			int d = sc.nextInt();
			if(!merge(l,r,d)) {//マージに成功するかどうか
				System.out.println("NO");
				return;
			}
		}
		System.out.println("Yes");
	}
	public static boolean merge(int l, int r, int d) {
		int lb = updateBelongs(l);//親の所属をだす
		int rb = updateBelongs(r);
		if(lb == rb) {//２点の親同士の所属が同じか
			return states[r] == states[l] + d;//同じなら満たしているかどうか
		} else {
			belongs[rb] = lb;//所属が違うなら右の人の親の所属を左の親と合わせる
			states[rb] = states[l] + d - states[r];//右の人の親の距離を変える
			return true;
		}
	}
	public static int updateBelongs(int a) {
		if(a == belongs[a]) {//aさんの所属番号がaかどうか(所属番号の親かどうか
			return a;
		} else {
			int temp = belongs[a];
			belongs[a] = updateBelongs(belongs[a]);//親を辿って親の所属番号にする
			states[a] += states[temp];//距離を更新
			return belongs[a];
		}
	}
}