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.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int M = scanner.nextInt();

    UnionFind uf = new UnionFind(N);

    for (int i = 0; i < M; i++) {
      int left = scanner.nextInt() - 1;
      int right = scanner.nextInt() - 1;
      int dist = scanner.nextInt();

      int[] lr = uf.find(left);
      int[] rr = uf.find(right);

      if (lr[0] == rr[0]) {
        if (rr[1] - lr[1] != dist) {
          System.out.println("No");
          return;
        }
      } else {
        uf.union(left, right, dist);
      }
    }

    System.out.println("Yes");
  }

  static class UnionFind {
    private final int[] table;
    private final int[] delta;
    private final int[] ranks;

    public UnionFind(int size) {
      this.table = new int[size];
      this.delta = new int[size];
      this.ranks = new int[size];
      for (int i = 0; i < size; i++) {
        table[i] = i;
      }
    }

    public int[] find(int x) {
      if (table[x] != x) {
        int[] res = find(table[x]);
        table[x] = res[0];
        delta[x] += res[1];
      }
      return new int[] {table[x], delta[x]};
    }

    public void union(int x, int y, int dist) {
      int[] xRoot = find(x);
      int[] yRoot = find(y);
      if (xRoot[0] == yRoot[0]) return;

      if (ranks[xRoot[0]] < ranks[yRoot[0]]) {
        int[] tmp = xRoot;
        xRoot = yRoot;
        yRoot = tmp;
        dist = -dist;
      }

      table[yRoot[0]] = xRoot[0];
      delta[yRoot[0]] = xRoot[1] + dist - yRoot[1];
      if (ranks[xRoot[0]] == ranks[yRoot[0]]) ranks[xRoot[0]]++;
    }
  }
}