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.ArrayList;
import java.util.Arrays;
import java.util.List;
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 l[] = new int[m];
        int r[] = new int[m];
        int d[] = new int[m];
        List<Info>[] list = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            list[i] = new ArrayList<Info>();
        }
        for (int i = 0; i < m; i++) {
            l[i] = sc.nextInt();
            r[i] = sc.nextInt();
            d[i] = sc.nextInt();
            Info ri = new Info(r[i] - 1, d[i]);
            Info le = new Info(l[i] - 1, -d[i]);
            list[l[i] - 1].add(ri);
            list[r[i] - 1].add(le);
        }
        int x[] = new int[n];
        Arrays.fill(x, Integer.MAX_VALUE);
        boolean ans = true;
        for (int i = 0; i < n; i++) {
            if (x[i] == Integer.MAX_VALUE) {
                x[i] = 0;
                ans &= DFS(i, x, list);

            }
        }
        if (ans) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

    static boolean DFS(int i, int[] x, List<Info>[] lists) {
        List<Info> xlist = lists[i];
        for (int j = 0; j < xlist.size(); j++) {
            int to = xlist.get(j).to;
            int cost = xlist.get(j).cost;
            if (x[to] == Integer.MAX_VALUE) {
                x[to] = x[i] + cost;
                if (!DFS(j, x, lists)) {
                    return false;
                }
            } else {
                if (x[to] != x[i] + cost) {
                    return false;
                }
            }

        }
        return true;
    }
}

class Info {
    int to;
    int cost;

    Info(int to, int cost) {
        this.to = to;
        this.cost = cost;
    }
}