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 Python (3.4.3) (judge verdict: Wrong Answer)

# coding:utf-8
from pprint import pprint

def bellman_ford(V, E, s):
    """ 始点sから残りのノードへの最短経路を求めるベルマンフォード法
    :param V: 頂点数
    :param E: (ノード1, ノード2, コスト)のタプルのリスト
    :param s: 始点とするノード
    :return : (距離リスト, 最短経路, 負の辺によるループが存在するか)a
    """
    ##### 前処理 #####
    from collections import defaultdict
    # INFの定義
    INF = 2**63 - 1
    # 最短経路情報を格納する辞書
    dist = defaultdict(lambda: INF) # 始点から始点への移動コスト
    dist[s] = 0
    # 移動経路の履歴
    pre = {}

    ##### 最短経路算出 #####
    # 高々|V|-1回の探索でいい(|V|回以上更新がある場合は負の閉路が存在)
    for i in range(V-1):
        for n1, n2, w in E:
            # 始点からi回の探索でたどり着けない頂点は考慮しない
            # 始点 -> n2 までの最短距離 > 始点 -> n1 までの最短距離 + n1 -> n2 への移動コスト
            # の場合より短い経路が発見できたと言うことなので情報を更新
            if dist[n1] != INF and dist[n2] > dist[n1] + w:
                dist[n2] = dist[n1] + w
                pre[n2] = n1

    ##### 負の閉路検出 #####

    is_cycle = False

    for n1, n2, w in E:
        # 正しい最短距離が求まったと仮定(重要)
        # 始点 -> n1 までの最短距離 + n1 -> n2 への移動コスト < 始点 -> n2 までの最短経路
        if dist[n1] + w < dist[n2]:
            is_cycle = True

    return (dist, pre, is_cycle)




n, m = map(int, input().split(' '))
# 辺の情報
e = []
for _ in range(m):
    a, b, c = map(int, input().split(' '))
    e.append((a, b, -c))


dist, pre, is_cycle = bellman_ford(n, e, 1)

'''
print()
print('##### GRAPH2 (負の閉路あり) #####')
print()
print('===== dist =====')
pprint(dist)
print()
print('===== pre =====')
pprint(pre)
print()
print('===== is_cycle =====')
print(is_cycle)
print('#################################')
print()
'''



if is_cycle:
    g = n
    for i in range(n):
        g = pre[g]
        if g == 1:
            break

    if g == 1:
        print(-dist[n])
    else:
        print('inf')

else:
    print(-dist[n])