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# include <bits/stdc++.h>
using namespace std ;
const int N = 30 ;
const int INF = 0x3f3f3f3f ;
int a [ N ] ;
bool st [ N ] ;
int n , k ;
int cnt ;
vector < int > path ;
bool isPrime ( int n ) {
if ( n < 2 ) return false ;
for ( int i = 2 ; i < = n / i ; i + + )
if ( n % i = = 0 ) return false ;
return true ;
}
/**
* 功能:遍历每个箱子,向箱子里装卡片
* @param step 走在第几个箱子面前
* @param sum 已经获得的数字和
* @param start 从哪个数字可以开始进行选择
* 举个栗子：
* 3+7+19
* 3+19+7
* 7+3+19
* 7+19+3
* 19+3+7
* 19+7+3 这样就出来6个， ， ！
* 方法就是“从上一次选择的数字后面一位进行再次选择！”
* @return 获取可行方法数
*/
void dfs ( int step , int sum , int start ) {
//当走到第k+1个虚拟箱子面前，
if ( step = = k + 1 ) {
if ( isPrime ( sum ) ) {
cnt + + ;
for ( int i = 0 ; i < path . size ( ) ; i + + )
cout < < path [ i ] < < " " ;
cout < < endl ;
}
return ;
}
//在没有走完箱子时，考虑当前箱子放哪张卡片
for ( int i = start ; i < = n ; i + + ) {
if ( ! st [ i ] ) {
st [ i ] = true ;
path . push_back ( i ) ;
dfs ( step + 1 , sum + a [ i ] , i + 1 ) ;
path . pop_back ( ) ;
st [ i ] = false ;
}
}
}
/*
4 3
3 7 12 19
答案：
*/
int main ( ) {
cin > > n > > k ;
for ( int i = 1 ; i < = n ; i + + ) cin > > a [ i ] ;
dfs ( 1 , 0 , 1 ) ;
cout < < cnt < < endl ;
return 0 ;
}
