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: Accepted)

import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		ArrayList<Edge>[] list = new ArrayList[n+1];
		for(int i = 1; i <= n; i++){
		    list[i] = new ArrayList<Edge>();
		}
		for(int i = 0; i < m; i++){
		    int l = sc.nextInt();
		    int r = sc.nextInt();
		    int d = sc.nextInt();
		    list[l].add(new Edge(l, r, d));
		    list[r].add(new Edge(r, l, -d));
		}

		// 計算
		String result = "Yes";
		int[] dist = new int[n+1];
		Arrays.fill(dist, Integer.MAX_VALUE);
		int[] visited = new int[n+1];
		for(int i = 1; i < n+1; i++){
		    if(visited[i] == 1) continue;
		    PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
		    dist[i] = 0;
		    pq.add(i);
		    while(!pq.isEmpty()){
		        int now = pq.poll();
		        if(visited[now] == 1) continue;
		        visited[now] = 1;
		        for(Edge e : list[now]){
		            int from = e.from;
		            int to = e.to;
		            int d = e.weight;
		            if(dist[to] == Integer.MAX_VALUE){
		                dist[to] = dist[from] + d;
		            }else{
		                if(dist[to] != dist[from] + d) result = "No";
		            }
		            pq.add(to);
		        }
		    }
		}
		
		// 出力
		System.out.println(result);
	}
}

class Edge {
    int from;
    int to;
    int weight;
    
    Edge(int from, int to, int weight){
        this.from = from;
        this.to = to;
        this.weight = weight;
    }

    public String toString(){
        return "from:" + from + " to:" + to + " weight:" + weight;
    }   
}