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: Runtime Error)

#include <stdio.h>
long long 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 main(void){
    int n;

    int i, j, k;
    scanf("%d", &n);
    for(i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    
    
    for(i = 0; i < n; i++){
        b[i] = 0;//0:not used 1:used
    }
    int cnt;
    
    int ans = 0;
    while(1){
        
        ans++;
        // printf("ans = %d \n", ans);
        //listの先頭を取得し代入
        cnt = 0;
        while(1){
            if(b[cnt] == 0){
                list[0] = a[cnt];
                break;
            }
            cnt++;
        }
        
        j = 0;//list配列において，一番後ろの要素が入っている
        for(i = cnt + 1; i < n; i++){
            if(b[i] == 1){//使用済だった場合はスキップ
                continue;
            }
            if(list[j] < a[i]){
                j++;
                list[j] = a[i];
            }else if(list[j] == a[i]){
                continue;
            }else if(list[0] >= a[i]){
                list[0] = a[i];
                continue;
            }else{
                int l = 0;
                int r = j;
                while(r - l != 1){
                    int mid = (r + l)/2;
                    if(list[mid] < a[i]){
                        l = mid;
                    }else if(list[mid] > a[i]){
                        r = mid;
                    }else if(list[mid] == a[i]){
                        r--;
                    }
                }
                list[r] = a[i];
            }
        }


        cnt = 0;
        for(i = 0; i <= j; i++){
            while(1){
                if(b[cnt] == 1){
                    cnt++;
                    continue;
                }
                if(list[i] == a[cnt]){
                    b[cnt] = 1;
                    cnt++;
                    break;
                }
                cnt++;
            }
        }

        //debug
        // for(i = 0; i < n; i++){
        //     printf("%d ", b[i]);
        // }
        // printf("\n");

        int flag = 0;
        for(i = 0; i < n; i++){
            if(b[i] == 0){//未使用の要素がまだあれば続行
                flag = 1;
                break;
            }
        }

        if(flag == 0){
            break;
        }
    }
    printf("%d\n", ans);
}