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 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <limits.h>	
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define EPS 1e-10
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r);
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r);
#define TYPE int
#define MEMSET(a) memset(a,0,sizeof(a))
const int mod=(int)1e09+7;

int in(void){
	int i;scanf("%d",&i);
	return i;
}
long long llin(void){
	long long i;scanf("%lld",&i);
	return i;
}
double din(void){
	double i;scanf("%lf",&i);
	return i;
}
void chin(char s[]){
	scanf("%s",s);
}
void print(int a){
	printf("%d\n",a);
}
void llprint(long long a){
	printf("%lld\n",a);
}
void dprint(double a){
	printf("%.10f\n",a);
}
void print2(int a,int b){
	printf("%d %d\n",a,b);
}
long long max(long long a,long long b){
	return a>b?a:b;
}
long long min(long long a,long long b){
	return a<b?a:b;
}
long long llabs(long long a){
	return a>0?a:-a;
}
double dmax(double a,double b){
	return a>b?a:b;
}
int cmp(const void *a,const void *b){
	return *(TYPE *)a-*(TYPE *)b;
}
int cmp_r(const void *a,const void *b){
	return *(TYPE *)b-*(TYPE *)a;
}
int char_cmp(const void *a,const void *b){
	return strcmp((char *)a,(char *)b);
}
int char_cmp_r(const void *a,const void *b){
	return strcmp((char *)b,(char *)a);
}
void swap(int *a,int *b){
	int t=*a;
	*a=*b;
	*b=t;
}
long long gcd(long long x,long long y){
	return x%y?gcd(y,x%y):y;
}
long long lcm(long long x,long long y){
	return x/gcd(x,y)*y;
}

// konoshita kara kaku
#define N 1000
#define M 2000
int a[M],b[M];
long long c[M];
int n,m;

long long bellman_ford(void){
	long long dist[N];
	int i,j;
	rep(i,n)dist[i]=INF;
	dist[0]=0;
	
	rep(j,n-1){
		rep(i,m){
			if(dist[a[i]-1]==INF)continue;
			if(dist[b[i]-1]>dist[a[i]-1]+c[i])
				dist[b[i]-1]=dist[a[i]-1]+c[i];
		}
	}
	
	long long ans = dist[n-1];
	
	int negative[N]={};
	
	rep(j,n){
		rep(i,m){
			if(dist[a[i]-1]==INF)continue;
			if(dist[b[i]-1]>dist[a[i]-1]+c[i]){
				dist[b[i]-1]=dist[a[i]-1]+c[i];
				negative[b[i]-1]=1;
			}
			if(negative[a[i]-1]==1)negative[b[i]-1]=1;
		}
	}
	
	return negative[n-1]?INF:ans;
}

int main(void){
	int i;
	n=in(),m=in();
	rep(i,m){
		a[i]=in();
		b[i]=in();
		c[i]=llin();
		c[i]*=-1;
	}
	long long ans=bellman_ford();
	if(ans==INF)puts("inf");
	else llprint(-ans);
	return 0;
}