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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include<bitset>

using namespace std;

#define M 1000000007
#define pb push_back
#define f first
#define s second
#define rep(i, st, ed) for(int i=st; i<ed; i++)
#define repn(i, st, ed) for(int i=st; i<=ed; i++)
#define repb(i, ed, st) for(int i=ed; i>=st; i--)

typedef long long ll;
typedef unsigned long long int llt;
typedef long double ld;

const long double PI  =3.141592653589793238463;
const int N = 5e5 + 10;

vector<int> g[N];
bool viss[N];
int vis[N];

// Normal DFS
void dfs(int curr, int par)
{
  for(auto x: g[curr])
    if(x != par)
      dfs(x, curr);
}

// DFS with visiting status
void dfs1(int curr, int par)
{
  viss[curr] = 1;
  for(auto x: g[curr])
    if(!viss[x])
      dfs1(x, curr);
}

// DFS with strongly connected component


// XOR from 0 to n
ll fxor(ll n)
{
  if(n < 0) return 0;
  if(n % 4 == 3) return 0;
  else if(n % 4 == 0) return n;
  else if(n % 4 == 1) return n^(n-1);
  else if(n%4 == 2) return n^(n-1)^(n-2);
}

// Power in logn with mod 1000000007
llt ppow(llt x, llt y)
{
  llt res = 1;
  while(y > 0){
    if(y&1)
      res = (res * x)%M;
    y = y >> 1;
    x = (x * x)%M;
  }
  return res;
}

// Factorial with dp
llt factdp[N];
llt fact(llt x)
{
  if(x == 0) return 1;
  if(x == 1) return 1;
  if(factdp[x] != -1) return factdp[x];
  return factdp[x] = (x*fact(x-1))%M;
}

// String return to base b
string ansConv ="";
void convert10tob(llt N, int b)
{
     if (N == 0)
        return;
     int x = N % b;
     N /= b;
     if (x < 0)
        N += 1;
     convert10tob(N, b);
     ansConv += to_string(x);
     return;
}

// Convert string to num wiht base b
llt convertstrtob(string a, int b)
{
  llt aa = 0;
  llt mul = 1;
  for(int i=a.length()-1; i>=0; i--)
  {
    int curr = a[i] - '0';
    aa += (curr * mul);
    mul *= b;
  }
  return aa;
}

// Union Find algo with array
// Find()
int findpar(int curr, int arr[])
{
  while(arr[curr] != curr) curr = arr[curr];
  return curr;
}
// Union()
void Union(int b, int spa, int arr[])
{
  while(arr[b] != b)
  {
    int c = arr[b];
    arr[b] = spa;
    b = c;
  }
  arr[b] = spa;
}

// Main Code
vector<pair<ll, ll>> gp[1005];
int n , m;
ll dist[1005];
bool cyc = false;

void dfscyc(ll curr)
{
  viss[curr] = 1;
  for(auto x: gp[curr])
  {
    ll nx = x.f;
    ll w = x.s;
    if(dist[nx] < (dist[curr]+w))
    {
      if(viss[nx] == 1)
      {
        cyc = true;
        return;
      }
      else
      {
        dist[nx] = dist[curr] + w;
        dfscyc(nx);
      }
    }
  }
  viss[curr] = 0;
}

int main()
{

  cin>>n>>m;
  repn(i, 0, n)
    dist[i] = LONG_MIN;
  dist[1] = 0;
  rep(i, 0, m)
  {
    int a, b;
    ll c;
    cin>>a>>b>>c;
    gp[a].pb({b, c});
  }
  dfscyc(1);
  if(cyc) puts("inf");
  else
  {
    cout<<dist[n]<<endl;
  }
  return 0;
}