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: Time Limit Exceeded)

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();

      Result lr = uf.find(left);
      Result rr = uf.find(right);

      if (lr.root == rr.root) {
        if (rr.dist - lr.dist != 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;

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

    Result find(int x) {
      if (table[x] != x) {
        Result parentRes = find(table[x]);
        return new Result(parentRes.root, parentRes.dist + delta[x]);
      }
      return new Result(x, 0);
    }

    void union(int x, int y, int dist) {
      Result xRoot = find(x);
      Result yRoot = find(y);
      if (xRoot.root == yRoot.root) return;

      table[yRoot.root] = xRoot.root;
      delta[yRoot.root] = xRoot.dist + dist - yRoot.dist;
    }
  }

  static class Result {
    public final int root;
    public final int dist;

    Result(int root, int dist) {
      this.root = root;
      this.dist = dist;
    }
  }
}