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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

class P
{
    static void Main()
    {
        int[] NM = Console.ReadLine().Split().Select(int.Parse).ToArray();
        Tree tree = new Tree(NM[0]);
        for (int i = 0; i < NM[1]; i++)
        {
            int[] LRD = Console.ReadLine().Split().Select(int.Parse).ToArray();
            if(!tree.Union(LRD[0] - 1, LRD[1] - 1, LRD[2]))
            {
                Console.WriteLine("No");
                return;
            }
        }
        Console.WriteLine("Yes");
    }
}

public class Tree
{
    //親、データ、加算要素(下からたどってってごにょごにょする)
    TreeD[] data;
    public Tree(int count)
    {
        data = Enumerable.Range(0, count).Select(x => new TreeD(x, 0)).ToArray();
    }
    public bool Union(int x, int y, int Diff)
    {
        int xparent = Parent(x);
        int yparent = Parent(y);
        int xvalue = Value(x);
        int yvalue = Value(y);
        if (xparent != yparent)
        {
            data[yparent].Parent = xparent;
            data[yparent].Value = Diff + (xvalue - yvalue);
        }
        else
        {
            if (yvalue - xvalue != Diff) return false;
        }
        return true;
    }
    public int Parent(int x)
    {
        form(x);
        return data[x].Parent;
    }
    public int Value(int x)
    {
        form(x);
        return data[x].Value;
    }
    //強制的にひっつかせる奴
    private void form(int x)
    {
        while (data[data[x].Parent].Parent != data[x].Parent)
        {
            //親のAddを足して親の親にひっつく
            data[x].Value += data[data[x].Parent].Value;
            data[x].Parent = data[data[x].Parent].Parent;
        }
    }
}
class TreeD
{
    public int Parent;
    public int Value;
    public TreeD(int p, int a)
    {
        Parent = p;
        Value = a;
    }
}