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# include <bits/stdc++.h>
using namespace std ;
const double eps = 1e-8 ;
const int INF = 0x3f3f3f3f ;
const int N = 1010 , M = 5010 ;
int n , m ;
int f [ N ] , cnt [ N ] ;
double dist [ N ] ;
bool st [ N ] ;
// 邻接表
int idx , h [ N ] , e [ M ] , w [ M ] , ne [ M ] ;
void add ( int a , int b , int c ) {
e [ idx ] = b , w [ idx ] = c , ne [ idx ] = h [ a ] , h [ a ] = idx + + ;
}
bool check ( double mid ) {
queue < int > q ;
memset ( cnt , 0 , sizeof cnt ) ;
memset ( dist , 0x3f , sizeof dist ) ;
memset ( st , false , sizeof st ) ;
for ( int i = 1 ; i < = n ; i + + ) {
q . push ( i ) ;
st [ i ] = 1 ;
}
while ( q . size ( ) ) {
int u = q . front ( ) ;
q . pop ( ) ;
st [ u ] = 0 ;
for ( int i = h [ u ] ; ~ i ; i = ne [ i ] ) {
int v = e [ i ] ;
// 最短路
if ( dist [ v ] > dist [ u ] + w [ i ] * mid - f [ u ] ) {
dist [ v ] = dist [ u ] + w [ i ] * mid - f [ u ] ;
// 判负环
cnt [ v ] = cnt [ u ] + 1 ;
if ( cnt [ v ] > = n ) return 1 ;
if ( ! st [ v ] ) {
q . push ( v ) ;
st [ v ] = 1 ;
}
}
}
}
return 0 ;
}
int main ( ) {
scanf ( " %d %d " , & n , & m ) ;
for ( int i = 1 ; i < = n ; i + + ) scanf ( " %d " , & f [ i ] ) ; // 每个点都有一个权值f[i]
// 初始化邻接表
memset ( h , - 1 , sizeof h ) ;
int a , b , c ;
while ( m - - ) {
scanf ( " %d %d %d " , & a , & b , & c ) ;
add ( a , b , c ) ;
}
// 浮点数二分
double l = 0 , r = INF ;
while ( r - l > eps ) {
double mid = ( l + r ) / 2 ;
if ( check ( mid ) )
l = mid ; // 存在负环时, mid再大一点,最终取得01分数规则的最大值
else
r = mid ; // 不存在负环时, mid再小一点
}
printf ( " %.2lf \n " , l ) ;
return 0 ;
}