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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;

typedef unsigned long long int ulint;
typedef long long int lint;

int main(){
  ulint N;
  cin >> N;

  vector<ulint> A;
  
  for (int i = 0; i < N; ++i) {
    ulint a;
    cin >> a;
    A.push_back(a);
  }

  vector<ulint> C(N, (ulint) -1);
  
  for (int i = N - 1; i >= 0; --i) {
    // TODO: binary search here.
    auto it = upper_bound(C.begin(), C.end(), A[i]);
    // int j = 0;
    // while (A[i] >= C[j]) {
    //   ++j;
    // }
    *it = A[i];
    // C[j] = A[i];    
  }

  // cout << "C= " << endl;
  // for (int i = 0; i < N; ++i) {
  //   cout << C[i] << " ";
  // }
  // cout << endl;
  
  auto it = find(C.begin(), C.end(), (ulint) -1);

  ulint val =  it - C.begin();

  cout << val << endl;

}