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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    a, b, c = LI()
    if a <= c <= b:
        print("Yes")
        return
    print("No")
    return

#B
def B():
    n, m = LI()
    e = defaultdict(int)
    for i in range(1, n + 1): e[i] = 0
    for i in range(m):
        a, b = LI()
        e[a] += 1
        e[b] += 1
    for i in e.values():
        print(i)
    return

#C
def C():
    n, k = LI()
    d = defaultdict(int)
    for i in range(1, 10**5+1): d[i] = 0
    for i in range(n):
        a, b = LI()
        d[a] += b
    for key, valus in d.items():
        k -= valus
        if k <= 0:
            print(key)
            return
    return

#D
def D():
    n, m = LI()
    edg = defaultdict(list)
    r_edg = defaultdict(list)
    for _ in range(m):
        a, b, c = LI_()
        edg[a].append((b, c + 1))
        r_edg[b].append(a)
    q = deque()
    q.append(n-1)
    check = [True] * n
    check[n-1] = False
    while q:
        now = q.pop()
        for e in r_edg[now]:
            if check[e]:
                check[e] = False
                q.appendleft(e)
    def dfs(v, score):
        res = False
        for e, c in edg[v]:
            if e == i and score > 0:
                return True
            if close[e]:
                close[e] = False
                res |= dfs(e, score + c)
                close[e] = True
        return res
    for i in range(n):
        if check[i]:
            continue
        close = [True] * n
        if dfs(i, 0):
            print("inf")
            return
    def dfsans(v, score):
        res = -inf
        for e, c in edg[v]:
            if e == n - 1:
                res = max(res, score + c)
                continue
            if check[e]:
                check[e] = False
                res = max(res, dfsans(e, score + c))
                check[e] = True
        return res
    check = [True] * n
    print(dfsans(0, 0))
    return

#Solve
if __name__ == '__main__':
    D()