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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>//random,NULL
#include <string.h>
#include <time.h>
#include <unistd.h>//sleep()
#include <math.h>

#define true 1
#define false 0

#define rep(i,n) for(int i=0; i<((int)n); i++)
#define REP(i,n) for(int i=0; i<=((int)n); i++)
#define srep(i,a,n) for(int i=a; i<((int)n); i++)
#define SREP(i,a,n) for(int i=a; i<=((int)n); i++)
#define rrep(i,n) for(int i=n-1; i>=0; i--)
#define RREP(i,n) for(int i=n; i>=0; i--)
#define INF 10000000000000

//有向グラフ
//cが負の値も取りうることと、N^3はデカすぎることより
//bellman_ford法を用いる。
//今回は普通のベルマンではないが。
/*
  アルゴリズム概要
  N-1回実行する間に、頂点Nの値は少なくとも１回更新される。
  まずはこのN-1回の中で最大化される時を見つける
  次に、N回目にも値が更新された時、閉路があるということなので、infになるかを調べる。
  次のN回の内（ループは最大でN）、さらに大きくなることがあれば、inf
*/

long bellman_ford(int N, int M, int **edge){
  long *d=(long *)malloc(sizeof(long)*N);
  rep(i,N){
    d[i]=INF;
  }
  d[0]=0;

  rep(k,N-1){
    //int update=false;
    rep(j,M){
      if(d[edge[j][0]]!=INF && (d[edge[j][1]]==INF || d[edge[j][0]]+edge[j][2]>d[edge[j][1]])){
        d[edge[j][1]]=d[edge[j][0]]+edge[j][2];
        //update=true;
      }
    }
  }
  // rep(i,N){
  //   printf("d[%d] = %d, ", i, d[i]);
  // }
  // printf("\n");
  //この時点で、d[N-1]にはN-1回の内で最大値が入っている。
  long record=d[N-1];

  rep(k,N){
    rep(j,M){
      if(d[edge[j][0]]+edge[j][2]>d[edge[j][1]]){
        d[edge[j][1]]=d[edge[j][0]]+edge[j][2];
      }
    }
  }
  // rep(i,N){
  //   printf("d[%d] = %d, ", i, d[i]);
  // }
  // printf("\n");

  if(record<d[N-1]){
    return INF;
  }
  else{
    return record;
  }
}




int main(){
  char buf[1024];
  int N,M;

  fgets(buf,sizeof(buf),stdin);
  sscanf(buf, "%d %d", &N, &M);

  int *array=(int *)malloc(sizeof(int)*N);
  rep(i,N){
    ;
  }

  int **edge=(int **)malloc(sizeof(int *)*M);
  rep(i,M){
    edge[i]=(int *)malloc(sizeof(int)*3);
  }
  rep(i,M){
    fgets(buf,sizeof(buf),stdin);
    sscanf(buf, "%d %d %d", &edge[i][0], &edge[i][1], &edge[i][2]);
    edge[i][0]--;
    edge[i][1]--;
  }

  long longest=bellman_ford(N, M, edge);

  if(longest==INF){
    printf("inf\n");
  }
  else{
    printf("%ld\n", longest);
  }

  return 0;
}