Problem:
Score : 400 points

Problem Statement

There is a directed graph with N vertices and M edges.
The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i.
We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0.
The player can move the piece as follows:

When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i.

The player can end the game only when the piece is placed at vertex N.
The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be?
If it is possible to increase the score indefinitely, print inf.

Constraints

2≤N≤1000

1≤M≤min(N(N-1),2000)

1≤a_i,b_i≤N (1≤i≤M)

a_i≠b_i (1≤i≤M)

a_i≠a_j or b_i≠b_j (1≤i<j≤M)

-10^9≤c_i≤10^9 (1≤i≤M)

c_i is an integer.

In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print inf.

Sample Input 1

3 3
1 2 4
2 3 3
1 3 5

Sample Output 1

7

There are two ways to move the piece to vertex N=3:

vertex 1 → vertex 2 → vertex 3 : score 4+3=7

vertex 1 → vertex 3 : score 5

Thus, the maximum possible score at the end of the game is 7.

Sample Input 2

2 2
1 2 1
2 1 1

Sample Output 2

inf

It is possible to increase the score indefinitely by alternating between vertex 1 and 2.

Sample Input 3

6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000

Sample Output 3

-5000000000

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)

import scala.collection.mutable.ArrayBuffer

object Main {

  val INF = Long.MaxValue
  case class Edge(to: Int, weight: Int)
  def main(args: Array[String]): Unit = {
    // 仮定, スコアが無限に増えるケースでは閉路が存在するから, inf
    // つまり, 正常な場合は閉路がないことから有効非巡回グラフになっているので
    // トポロジカルソート版ベルマンフォードが有効
    // 閉路検出は, トポロジカルソート順に再度緩和した時, 緩和できるかどうか?
    // なお, 1からVまで必ず到達できることを保証しているが, それ以外が連結かどうかを示していない
    val sc = new java.util.Scanner(System.in)

    val V, E = sc.nextInt()
    val G = Array.fill(V+1)(new ArrayBuffer[Edge]())
    for (_ <- 0 until E) {
      val u, v, w = sc.nextInt()
      G(u).append(Edge(v, -w))
    }

    bellmanFord(V, G, 1) match {
      case Some(v) => println(v)
      case _ => println("inf")
    }
  }

  def bellmanFord(V: Int, G: Array[ArrayBuffer[Edge]], s: Int): Option[Long] = {
    val dist = Array.fill(V+1)(INF)
    dist(s) = 0

    def relax(u: Int, v: Int, w: Int): Boolean =
      if (dist(u) != INF && dist(v) > dist(u) + w) {
        dist(v) = dist(u) + w
        true
      } else false

    val sorted = topologicalSort(V, G, 1)
    for {
      u <- sorted
      e <- G(u)
    } relax(u, e.to, e.weight)

    if (!sorted.exists(u => G(u).exists(e => relax(u, e.to, e.weight))))
      Some(-dist(V))
    else None
  }

  private def topologicalSort(V: Int, G: Array[ArrayBuffer[Edge]], s: Int): List[Int] = {
    val visited = new Array[Boolean](V+1)
    def dfs(stack: List[Int], sorted: List[Int]): List[Int] =
      if (stack.nonEmpty) {
        val u = stack.head
        if (u < 0) dfs(stack.tail, (-u)-1 :: sorted)
        else if (!visited(u)) {
          visited(u) = true
          dfs(
            G(u).foldLeft(-(u+1) :: stack.tail)(
              (stack, e) => e.to :: stack), sorted)
        } else dfs(stack.tail, sorted)
      } else sorted
    dfs(s :: Nil, Nil)
  }

}