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: Accepted)

/* bellman_ford.cpp
    ベルマンフォード法で、ある頂点から各頂点への最短距離を求めるアルゴリズム
    負の辺が含まれる場合も適用でき、負の閉路も検出可能

    始点から終点に行くまでに通らない負の閉路は検出したくない時:
        http://sigma1113.hatenablog.com/entry/2019/08/12/130042

    verified: AOJ GRL_1_B
        http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_B
*/

#include <bits/stdc++.h>
using namespace std;

struct Edge {
    long long from;
    long long to;
    long long cost;
};
using Edges = vector<Edge>;
const long long INF = 1LL << 60;

/* bellman_ford(Es,s,t,dis)
    入力: 全ての辺Es, 頂点数V, 開始点 s, 最短経路を記録するdis
    出力: 負の閉路が存在するなら ture
    計算量：O(|E||V|)
    副作用：dis が書き換えられる
*/
bool bellman_ford(const Edges &Es, int V, int s, vector<long long> &dis) {
    dis.resize(V, INF);
    dis[s] = 0;
    int cnt = 0;
    while (cnt < V) {
        bool end = true;
        for (auto e : Es) {
            if (dis[e.from] != INF && dis[e.from] + e.cost < dis[e.to]) {
                dis[e.to] = dis[e.from] + e.cost;
                end = false;
            }
        }
        if (end) break;
        cnt++;
    }
    cnt = 0;
    while (cnt < V) {
        bool end = true;
        for (auto e : Es) {
            if (dis[e.from] != INF && dis[e.from] + e.cost < dis[e.to]) {
                dis[e.to] = -INF;
                end = false;
            }
        }
        if (end) break;
        cnt++;
    }
    return dis[V - 1] == -INF;
}

int main() {
    int V, E;
    cin >> V >> E;
    Edges Es;
    for (int i = 0; i < E; i++) {
        long long s, t, d;
        cin >> s >> t >> d;
        s--, t--;
        Es.push_back({s, t, -(d)});
    }
    vector<long long> dis;
    if (bellman_ford(Es, V, 0, dis)) {
        cout << "inf" << endl;
    } else {
        cout << -dis[V - 1] << endl;
    }

    return 0;
}