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: Accepted)

#include<queue>
#include<stack>
#include<math.h>
#include<cmath>
#include<bitset>
#include<stdio.h>
#include<string>
#include<map>
#include<algorithm>
#include<vector>
#include<iostream>
#include<utility>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int main(){
  const int INF=2000000000;
  int n,i,dp[100010];
  scanf("%d\n",&n);
  vector<int> x(n);
  for(i=0;i<n;i++){
    scanf("%d\n",&x[i]);
  }
  reverse(x.begin(),x.end());
  for(i=0;i<=n;i++){
    dp[i]=INF;
  }
  for(i=0;i<n;i++){
    int min=-1,max=n,mid;
    while(abs(min-max)>1){
      mid=(min+max)/2;
      if(x[i]<dp[mid]){
        max=mid;
      }
      else{
        min=mid;
      }
    }
    dp[max]=x[i];
  }
  for(i=0;i<n;i++){
    if(dp[i]==INF){
      break;
    }
  }
  printf("%d\n",i);
  return 0;
}