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)

object Main {

  val INF = Long.MaxValue

  case class Edge(u: Int, v: Int, weight: Int)
  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val V, E = sc.nextInt()

    val edges = (0 until E).foldLeft(Nil: List[Edge])((lst, _) => {
      val u, v, w = sc.nextInt()
      Edge(u, v, -w) :: lst
    })

    // N頂点,M辺の重み付き有向グラフ
    // このグラフと駒を使用する
    // 最初, 駒を頂点1に置いて, プレイヤーのスコアを0とします
    // プレイヤーは次の条件で駒を繰り返し移動させることができます
    // * 頂点a_iに駒がある時, i番目の辺を利用して, 頂点b_iに移動する
    //   移動後にプレイヤーのスコアがc_i加算される
    // 頂点Nに駒がある時のみ, ゲームを終了できます.
    // なお, 与えられる有向グラフの上で、頂点1から頂点Nに移動できることは保証されています
    // プレイヤーがゲーム終了時のスコアをできるだけ大きくするような行動を取った時
    // ゲーム終了時のスコアは幾つになるでしょうか?

    // なんで, 閉路がある場合には, いくらでも回ればいいので, INFになる
    // なるべくスコアを大きくしたいので, この場合, c' = -c となる新しい重みを
    // 導入すると, u -> vに行くときになるべく大きい値を取るようになる(最短路なので)

    // 始点と終点が決まっているので, 単一始点最短経路問題に帰着する.
    // 負辺があるのでダイクストラは使えない.
    // 普通にO(VE)でもV = 1000, E = 2000 なので間に合う感じ
    bellmanFord(V, edges, 1) match {
      case Some(d) => println(-d(V))
      case _ => println("inf")
    }
  }

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

    dist(s) = 0L
    for {
      _ <- 0 until V - 1
      edge <- edges
    } relax(edge.u, edge.v, edge.weight)

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

    if (edges.forall(edge => !relax(edge.u, edge.v, edge.weight)))
      Some(dist)
    else None
  }

}