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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Diagnostics;

class Program
{
    static void Main(string[] args)
    {
        string[] cin = Console.ReadLine().Split();
        int n = int.Parse(cin[0]);
        int m = int.Parse(cin[1]);

        // 始点からの距離dist（初期値は無限。後でdist[始点]だけ0にする）
        // 配列やListは参照型なので普通に引数にした場合(値渡し)
        // 引数にしたdistは戻り値にしなくても結果を受け取れる
        long[] dist = new long[n];
        for (int i = 0; i < n; i++)
            dist[i] = long.MaxValue;

        // Bellman_ford(頂点数,始点,辺の数,dist)
        // 戻り値はbool[n]で始点→各頂点の途中に負閉路が存在するか(存在すればtrue)
        int s = 0; // 今回の始点は固定で0
        bool[] negative = Bellman_ford(n, s, m, dist);

        // 今回の目的地は固定でn-1
        // n-1までの経路に負閉路が存在すれば答えはinf
        // また、本問題ではコストの正負を反転させてるので元に戻す
        if (negative[n - 1]) Console.WriteLine("inf");
        else Console.WriteLine(-dist[n - 1]);
    }

    static bool[] Bellman_ford(int n, int s, int m, long[] dist)
    {
        // グラフの受け取りはこっちでやる
        // 頂点a→頂点b(コストc)
        string[] cin;
        int[] a = new int[m];
        int[] b = new int[m];
        long[] c = new long[m];
        for (long i = 0; i < m; i++)
        {
            cin = Console.ReadLine().Split();
            // 本問題では
            // ・入力のインデックスが1スタートなので0スタートに変更
            // ・コストの正負を反転させる
            a[i] = int.Parse(cin[0]) - 1;
            b[i] = int.Parse(cin[1]) - 1;
            c[i] = -long.Parse(cin[2]);

            //a[i] = int.Parse(cin[0]);
            //b[i] = int.Parse(cin[1]);
            //c[i] = long.Parse(cin[2]);
        }

        // 始点から始点の距離は0
        dist[s] = 0;

        // 第一ステップ：最短距離の算出
        // 始点を除くn-1個の頂点と全ての辺について貪欲的に最短距離を算出
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                // dist[a[j]]がlong.MaxValueだと桁あふれするのでその場合を除外
                if (dist[a[j]] != long.MaxValue && dist[b[j]] > dist[a[j]] + c[j])
                {
                    dist[b[j]] = dist[a[j]] + c[j];
                }
            }
        }

        // 第二ステップでdistの値が変化する場合、それは最短距離では無いので
        // 作業用にdist2作ったりする必要はない

        // 第二ステップ：負閉路の検出
        // 負閉路が無い場合はここまでで最短距離が出てるはず
        // 全ての辺について再計算して最短距離が変わるならそれは負閉路の存在を示す
        bool[] negative = new bool[n];

        for (int j = 0; j < m; j++)
        {
            if (dist[a[j]] != long.MaxValue && dist[b[j]] > dist[a[j]] + c[j])
            {
                dist[b[j]] = dist[a[j]] + c[j];
                negative[b[j]] = true;
            }
            // 負閉路を経由してきたやつはその先も負閉路経由となるので
            if (negative[a[j]]) negative[b[j]] = true;
        }
        return negative;
    }
}