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: Time Limit Exceeded)

#include<bits/stdc++.h>				//edit by thidailoc
using namespace std;				//team three wolves
typedef long long ll;
typedef vector<ll> vi;
#define fu(i,a,b)   for(ll i=a;i<b;i++)
#define fd(i,b,a)   for(ll i=b;i>=a;i--)
#define pb(i)	    push_back(i)
#define popb()    	pop_back()
#define sc(a) 	    scanf("%lld",&a)
#define sc2(a,b)    scanf("%lld%lld",&a,&b)
#define pr(a)	    printf("%lld ",a)
#define pr2(a,b)    printf("%lld %lld",a,b)
#define lowb(a,n,x) lower_bound(a,a+n,x) -a
#define lowb2(a,x)	lower_bound(a.begin(),a.end(),x) -a.begin()
#define all(x) 		(x).begin(), (x).end()
#define sz(a) 		(ll)a.size()
#define le			length()						
#define re 			return 	
#define debug(x)    cerr << #x << " = " << x << endl
#define INPUT       freopen("locin.txt", "r", stdin)
#define OUTPUT      freopen("locout.txt", "w", stdout)

double const eps = 1e-6;	//dung trong sai so
ll const Base=1e9+7;
vi dske[1005];
ll W[1005][1005],Onstack[1005],markcycle[1005],dp[1005],n;
class Info{
	public: ll v,w,prevv;
};
class cmp_w{
	public: bool operator()(Info a,Info b){
		return a.w>b.w;
	}
};
vi path;
bool CheckCycle(ll cur){

	fu(i,0,sz(dske[cur])){
		if(dske[cur][i]==n){
			fd(j,path.size()-1,0){
				if(markcycle[path[j]]==-1)re 1;
				if(path[j]==dske[cur][i])break;
			}
			continue;
		}
	//	debug(i);
		if(Onstack[dske[cur][i]]){
			ll sum=0,next_v=dske[cur][i];
			fd(j,path.size()-1,0){
				//markcycle[path[j]]=1;
				sum+=W[path[j]][next_v];
				next_v=path[j];
				if(path[j]==dske[cur][i])break;
			}
			
			fd(j,path.size()-1,0){
				//markcycle[path[j]]=1;
				markcycle[path[j]]=(sum==0?1:-1);
				if(path[j]==dske[cur][i])break;
			}
			continue;
		}
		Onstack[dske[cur][i]]=1;
		path.pb(dske[cur][i]);
		if(CheckCycle(dske[cur][i]))re 1;
		path.popb();
		Onstack[dske[cur][i]]=0;
	}
	re 0;
}
ll Dijkstra(ll start,ll target){
	priority_queue<Info,vector<Info>,cmp_w> pq;
	Info cur,x;
	ll xx,flag=0;;
	cur.v=start; cur.w=0;
	pq.push(cur);
	
	while(pq.size()){
		cur=pq.top();
		pq.pop();
		if(markcycle[cur.v]==-1)continue;
		if(dp[cur.v]<=cur.w)continue;
		flag=0;
		while(xx!=1){
			if(xx==cur.v){
				flag=1;break;
			}
			xx=x.prevv;
		}
		if(flag)continue;
		dp[cur.v]=cur.w;
		if(cur.v==target)continue;
	//	debug(cur.v);
		fu(i,0,sz(dske[cur.v])){
			if(dp[dske[cur.v][i]]>cur.w+W[cur.v][dske[cur.v][i]]){
				x.v=dske[cur.v][i];
				x.w=cur.w+W[cur.v][dske[cur.v][i]];
				x.prevv=cur.v;
				pq.push(x);
			}
		}
	}
	re -dp[n];
}
ll Solves(){
//	INPUT;
	ll m,k,cnt=0,ans=0,x,a,b,c;
	// PROCESSING IN HERE
	sc2(n,m);
	while(m--){
		sc2(a,b);
		dske[a].pb(b);
		sc(W[a][b]);
		W[a][b]=-W[a][b];
		//debug(W[a][b]);
	}
	
	Onstack[1]=Onstack[n]=1;
	fu(i,0,1005)dp[i]=1e15;
	//dp[1]=-1;
	path.pb(1);
	//if(CheckCycle(1))re cout<<"inf",0;
	
	Dijkstra(1,n);
	if(dp[n]==1e15)cout<<"inf";
	else cout<<-dp[n];
	re 0;
}


int main(){
	Solves();
}