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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
import java.math.*
import java.lang.Math.*

val readString: ()->String = {readLine()!!}
val readInt: ()->Int = {readLine()!!.toInt()}
val readLong: ()->Long = {readLine()!!.toLong()}
val readIntArray: ()->IntArray = {readLine()!!.split(' ').map{it.toInt()}.toIntArray()}
val readLongArray: ()->LongArray = {readLine()!!.split(' ').map{it.toLong()}.toLongArray()}
val errPrintln: (String)->Unit = {msg -> System.err.println(msg)}
val MOD = 1e9.toLong()+7
val INF = Int.MAX_VALUE/2
val LINF = Long.MAX_VALUE/2
data class IntPair(val first: Int, val second: Int)
data class To(val idx: Int, val cost: Long)


class Bellman(val eg: Array<ArrayList<To>>, val start: Int) {
    private var dist = LongArray(0)

    var hasCycle = false
    private set

    init {
        val N = eg.size

        dist = LongArray(N){-LINF}
        dist[start] = 0L

        for (i in 0 until N) {
            var updating = false
            for (x in 0 until N)  {
                for (t in eg[x]) {
                    val y = t.idx

                    if (dist[x] + t.cost > dist[y]) {
                        dist[y] = dist[x] + t.cost
                        updating = true
                    }
                }
            }
            if (!updating) break
        }

        hasCycle = calcCycle()
    }


    private fun calcCycle(): Boolean {
        val N = eg.size
        val ndst = dist.clone()
        //for (i in 0 until N) {
            //var updating = false
            for (x in 0 until N)  {
                for (t in eg[x]) {
                    val y = t.idx
                    if (ndst[x] + t.cost > ndst[y]) {
                        ndst[y] = ndst[x] + t.cost
                        //updating = true
                    }
                }
            }
            //if (!updating) break
        //}

        //dist.forEachIndexed{ i, v ->
        //    if (v != ndst[i]) return true
        //}
        //return false
        return ndst[N-1] != dist[N-1]
    }


    fun distance(i: Int): Long {
        return if(hasCycle) LINF else dist[i]
    }
}


fun solve(eg: Array<ArrayList<To>>): Long {
    val N = eg.size

    val bm = Bellman(eg, 0)

    return bm.distance(N-1)
}


fun main(args: Array<String>) {
    val (N, M) = readIntArray()
    val eg = Array(N){ArrayList<To>()}
    for (i in 0 until M) {
        var (a, b, c) = readLongArray()
        a--;b--;
        eg[a.toInt()].add(To(b.toInt(), c))  // 有向グラフ
    }

    val ans = solve(eg)

    if (ans == LINF) println("inf")
    else println(ans)
}