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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import kotlin.system.exitProcess

fun dfs (v : Array<Int?>, g : List<MutableList<Pair<Int, Int>>>, c : Int) : Boolean {
    g[c].forEach {
        if (v[it.first] == null) {
            v[it.first] = v[c]!! + it.second
            dfs(v, g, it.first)
        } else if (v[c]!! + it.second != v[it.first]) {
            return false
        }
    }

    return true
}

fun main (args : Array<String>) {
    var (n, m) = readLine()!!.split(" ").map{ it.toInt() }

    var g = (0..n-1).map{ mutableListOf<Pair<Int, Int>>() }
    (0..m-1).forEach {
        var lrd = readLine()!!.split(" ").map{ it.toInt() }
        g[lrd[0] - 1].add(Pair(lrd[1] - 1, lrd[2]))
        g[lrd[1] - 1].add(Pair(lrd[0] - 1, -lrd[2]))
    }

    var v = arrayOfNulls<Int>(n)
    for (i in 0..n-1) {
        if (v[i] != null)
            continue

        v[i] = 0
        if (dfs(v, g, i) == false) {
            println("No")
            exitProcess(0)
        }
    }

    println("Yes")
}