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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include<stdlib.h>
int min(int a, int b){return a < b ? a : b;}
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main(void){
    int n;
    int list[100000], a[100000];
    int b[100000];//既に配列を使ったかどうかを記録する

    int i, j, k;
    scanf("%d", &n);
    for(i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }


    b[0] = a[n-1];
    int cnt = 1;
    int c_list[100000];
    int b_min, b_min_i;
    int mid;
    for(i = n - 2; i >= 0; i--){
        int tmp = 0;//塗る事の出来る色をカウント
        //a[i]を何かの色で塗りたい
        //bはソートされているため，b[j] > a[i]を満たす最小のb[j]をとるjの値を二分探索で求める
        //ない場合はflag=0
        if(b[cnt - 1] <= a[i]){
            b[cnt] = a[i];
            cnt++;
        }else if(b[0] > a[i]){
            j = 0;
            b[j] = a[i];
        }else{
            int r = cnt;
            int l = 0;
            while(r - l != 1){
                mid = (r + l) / 2;
                if(b[mid] < a[i]){
                    l = mid;
                }else if(b[mid] > a[i]){
                    r = mid;
                }else{
                    l = mid;
                }
            }
            j = r;
            b[j] = a[i];
        }

        qsort(b, cnt, sizeof(int), compare_int);
    }

    printf("%d\n", cnt);

    return 0;    
}