import java.util.*;

public class Main {
    static final int N = 1010;
    static int[] f = new int[N], a = new int[N];
    /**
    * 明确f的含义：f[i]表示以a[i]结尾的最长上升子序列的长度
    * f[i] = max(f[j] + 1);(j在1 ~ i - 1之间)注：还需要和1取max，
    * 有可能存在不进人j循环，如i等于1时或者a[i]为当前的最小值
    * 对于本题答案就是遍历f数组得到最大值
    */
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int n = Integer.parseInt(sc.nextLine());

        String[] s = sc.nextLine().split(" ");
        for (int i = 1; i <= n; i++) a[i] = Integer.parseInt(s[i - 1]);

        int res = 0;
        for (int i = 1; i <= n; i++) {
            f[i] = 1;
            for (int j = 1; j < i; j++)
                if (a[i] > a[j]) f[i] = Math.max(f[i], f[j] + 1);

            res = Math.max(res, f[i]);
        }

        System.out.println(res);
    }
}