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 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];
        for (int i = 0; i < M; ++i) {
            L[i] = sc.nextInt() - 1;
            R[i] = sc.nextInt() - 1;
            D[i] = sc.nextInt();
        }

        System.out.println(solve(N, L, R, D) ? "Yes" : "No");

        sc.close();
    }

    static boolean solve(int N, int[] L, int[] R, int[] D) {
        Parent[] parents = new Parent[N];

        for (int i = 0; i < L.length; ++i) {
            Root rootL = findRoot(parents, L[i]);
            Root rootR = findRoot(parents, R[i]);

            if (rootL.index == rootR.index) {
                if (rootL.totalDelta + D[i] != rootR.totalDelta) {
                    return false;
                }
            } else {
                parents[rootR.index] = new Parent(rootL.index, rootL.totalDelta + D[i] - rootR.totalDelta);
            }
        }

        return true;
    }

    static Root findRoot(Parent[] parents, int node) {
        int rootIndex = node;
        int totalDelta = 0;
        while (parents[rootIndex] != null) {
            totalDelta += parents[rootIndex].delta;
            rootIndex = parents[rootIndex].index;
        }
        Root result = new Root(rootIndex, totalDelta);

        int p = node;
        while (p != rootIndex) {
            int nextIndex = parents[p].index;
            int nextTotalDelta = totalDelta - parents[p].delta;

            parents[p] = new Parent(rootIndex, totalDelta);

            p = nextIndex;
            totalDelta = nextTotalDelta;
        }

        return result;
    }
}

class Parent {
    int index;
    int delta;

    Parent(int index, int delta) {
        this.index = index;
        this.delta = delta;
    }
}

class Root {
    int index;
    int totalDelta;

    Root(int index, int totalDelta) {
        this.index = index;
        this.totalDelta = totalDelta;
    }
}