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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

#define INF 1000000010

int binary_search(int x, int l, int r, int *array){
  while(l <= r){
    int m = (r + l - 1) / 2 + 1;
    if(array[m - 1] >= x && x >= array[m]) return m;
    else if(array[m] < x) r = m;
    else l = m;
  }
}

int main(){
  int n;
  scanf("%d", &n);
  
  int a[n];
  int dp[n];
  
  for(int i = 0; i < n; i++){
    scanf("%d", &a[i]);
    dp[i] = -1;
  }
  
  int right = 1;
  dp[0] = a[0];
  for(int i = 1; i < n; i++){
    if(dp[right - 1] >= a[i]){
      dp[right] = a[i];
      right++;
    }else if(dp[0] < a[i]){
      dp[0] = a[i];
    }else{
      dp[binary_search(a[i], 0, right, dp)] = a[i];
    }
  }
  
  int len;
  for(len = 0; len < n; len++) if(dp[len] == -1) break;
  
  printf("%d", len);
  
  return 0;
}