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// 실제로 그 수가 무엇인지를 구해야 함.

// 역추적

#include <iostream>

using namespace std;

int a[1000];

int d[1000];

int v[1000];

void go(int p){

// ? -> ? -> ,,, a[v[p]]-> a[p]

if (p == -1){

return;

}

// 가장 아래 단계부터 출력된다.

go(v(p));

cout << a[p] <<' ';

}

int main(){

int n;

cin >> n;

for(int i=0; i<n; i++){

cin >> a[i];

}

for(int i=0; i<n; i++){

d[i]= 1;

v[i]=-1;

for(int j=0;j<i;j++){

if(a[j]<a[i]&& d[i]<d[j]+1){

d[i] = d[j]+1;

v[i] = j;

}

int ans = d[0];

int p = 0;

for(int i=0;i<n;i++){

if(ans<d[i]){

ans = d[i];

p = i;

}

cout <<ans<<'\n';

go(p);

cout<<'\n';

return 0;

}

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