#include<iostream>
using namespace std;
//思路是先将s全排列到k，再计数count++ 到m，若到了m则输出此时的s 
int s[10002]={0},k[10002]={0},used[10002]={0};//s为全排列的数组，k为输入数组， 
long long count = 0,n,m,pan = 0;
void solve(int x){
	if(count - 1 == m)return ;
	if(x == n+1){
		int y = 1;
		for(int i = 1; i <=n; i++){
			if(s[i] != k[i] ) y = 0;
		}
		if(y == 1)pan = 1;
		if(pan == 1)count++;
		if(count -1 == m)for(int i = 1; i <= n; i++)cout<<s[i]<<" ";
		return;
	}
	for(int i = 1; i <= n; i++){
		
		if(!used[i]){
			used[i] = 1;
			s[x] = i;
			solve(x+1);
			s[x] = 0;
			used[i] = 0;
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i = 1; i <=n ; i++ )cin>>k[i];
	solve(1);
	return 0;
}