Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

4 5
1 2
1 3
3 2
2 4
3 4

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in Scala (2.11.7) (judge verdict: Runtime Error)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val V, E = sc.nextInt()
    val INF = Long.MinValue / 10
    val G = Array.fill(V, V)(INF)
    var D = Array.fill(V, V)(INF)
    (0 until E) foreach { _ =>
      val u, v = sc.nextInt()
      G(u-1)(v-1) = 1
      D(u-1)(v-1) = 1
    }
    (0 until V) foreach { i =>
      D(i)(i) = 0L
    }

    var k = 0
    while (k < V) {
      var u = 0
      val Dn = Array.fill(V, V)(INF)
      while (u < V) {
        var v = 0
        while (v < V) {
          if (D(u)(k) != INF && D(k)(v) != INF)
            Dn(u)(v) = D(u)(v) max (D(u)(k) + D(k)(v))
          else
            Dn(u)(v) = D(u)(v)
          v += 1
        }
        u += 1
      }
      D = Dn
      k += 1
    }
    val res = D.maxBy(_.max).max

    def isNegativeCycle(D: Array[Array[Long]]): Boolean =
      (0 until V) exists { i =>
        D(i)(i) < 0
      }

    if (isNegativeCycle(D))
      println("NEGATIVE CYCLE")
    else
      println(res)
  }

}