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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int N;
  cin >>N;
  vector<int> a(N),b(1);
  for(int i=0; i<N; i++)
    cin >>a.at(i);
  b.at(0)=a.at(0);
  int s,t;
  for(int i=1; i<N; i++){
    s=b.size()-1;
    if( a.at(i)  <=  b.at(s) ){
      b.push_back(  a.at(i)  );
    }
    if( a.at(i)  >  b.at(0) ){
      b.at(0) =a.at(i);
    }
    
    if((a.at(i)  <=  b.at(0))
       &&( a.at(i)  >  b.at(s) )){
         if(s>300){
           for(int j=0; j<s; j++){
             if(a.at(i)  >  b.at(min(200*j,s)) ){
               t=200*(j-1);
               break;
             }
           }
         }
         else
           t=0;
         for(int j=0; j<s; j++){
           if( a.at(i)  >  b.at(j+t) ){
             b.at(j+t) = a.at(i);
             break;
           }
         }
       }
  }
  cout << b.size();  
}