#include <iostream>
#include <fstream>
using namespace std;

#define INFINITY 1e20

void PrintSequence(int v[100001], int S[100001], int val, int poz);
ofstream g("scmax.out");
int main()
{
    /**
     * Read the numbers from `scmax.in`.
     */
    ifstream f("scmax.in");
    int v[100001];
    int n; // The length of numbers array.
    f >> n;
    for (int i = 0; i < n; i++)
        f >> v[i];
    f.close();

    /**
     * Solve the problem using dynamic programming.
     */
    // Declare an array in which we will save our partial results for
    // every step. Initialize it with a big number (INFINITY).
    int S[100001];
    for (int i = 0; i < n; i++)
        S[i] = INFINITY;

    // The solution for the first step is the first number itself.
    // So, the length of the longest non-decreasing sequence is 1;
    S[0] = 1;

    // Iterate over all items of our array of numbers (v), except the first one
    // for which we have already calculated the result.
    for (int i = 1; i < n; i++)
    {
        S[i] = 1;
        // Iterate over elements less than `i` and check if v[i] is a valid solution
        // for any previous sequences.
        for (int j = 0; j < i; j++)
            if(v[j] < v[i] && S[j] + 1 > S[i])
                S[i] = S[j] + 1;
    }

    /**
     * Print the length of the longest non-decreasing sequence
     * in `scmax.out'.
     */
    int max = 0;
    int poz = 0;
    for (int i = 0; i < n; i++)
        if (S[i] > max)
        {
            max = S[i];
            poz = i;
        }

    g << max << '\n';

    /**
     * Print the non-decreasing sequence.
     */
    PrintSequence(v, S, max, poz);
    return 0;
}

void PrintSequence(int v[100001], int S[100001], int val, int poz)
{
    for (int i = poz - 1; i >= 0; i--)
        if (S[i] + 1 == val && v[i] < v[poz])
        {
            PrintSequence(v, S, val - 1, i);
            break;
        }
    g << v[poz] << " ";
}