Problem:
Score : 500 points

Problem Statement

You are given a sequence with N integers: A = \{ A_1, A_2, \cdots, A_N \}.
For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:

If A_i and A_j (i < j) are painted with the same color, A_i < A_j.

Find the minimum number of colors required to satisfy the condition.

Constraints

1 \leq N \leq 10^5

0 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of colors required to satisfy the condition.

Sample Input 1

5
2
1
4
5
3

Sample Output 1

2

We can satisfy the condition with two colors by, for example, painting 2 and 3 red and painting 1, 4, and 5 blue.

Sample Input 2

4
0
0
0
0

Sample Output 2

4

We have to paint all the integers with distinct colors.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        try (Scanner in = new Scanner(System.in)) {
            int N = in.nextInt();
            int[] A = new int[N];
            for (int i = 0; i < N; i++) {
                A[i] = in.nextInt();
            }

            int numColors = 1;
            int[] colorToA = new int[N];
            colorToA[0] = -A[0];
            for (int i = 1; i < N; i++) {
                int maxAColor = -1;
                {
                    int maxA = -1;
                    for (int color = 0; color < numColors; color++) {
                        if (-colorToA[color] < A[i]) {
//                            if (-colorToA[color] > maxA) {
                            maxA = -colorToA[color];
                            maxAColor = color;
//                            }
                            break;
                        }
                    }
//                    Utils.debug(i, maxAColor);
                }

                if (maxAColor == -1) {
                    colorToA[numColors] = -A[i];
                    numColors++;
                } else {
                    colorToA[maxAColor] = -A[i];
                }
            }
            System.out.println(numColors);
        }
    }

    private static int lb(int[] a, int l, int r, int key) {
        int index = Arrays.binarySearch(a, l, r, key);
        while (index > 0 && a[index - 1] == a[index]) {
            index = Arrays.binarySearch(a, l, index - 2, key);
        }
        return index;
    }
}