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: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

void rsortr_u32(void *base, const int count){
	uint32_t *a=base;
	uint32_t *b=malloc(count*sizeof(uint32_t));
	int c[0x100];
	memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i]>>(0<<3))&0xFF]++; } for(int i=0xFE; i>=0; i--) c[i]+=c[i+1]; for(int i=count-1; i>=0; i--){ b[--c[(a[i]>>(0<<3))&0xFF]]=a[i]; }{void *t=a; a=b; b=t;}
	memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i]>>(1<<3))&0xFF]++; } for(int i=0xFE; i>=0; i--) c[i]+=c[i+1]; for(int i=count-1; i>=0; i--){ b[--c[(a[i]>>(1<<3))&0xFF]]=a[i]; }{void *t=a; a=b; b=t;}
	memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i]>>(2<<3))&0xFF]++; } for(int i=0xFE; i>=0; i--) c[i]+=c[i+1]; for(int i=count-1; i>=0; i--){ b[--c[(a[i]>>(2<<3))&0xFF]]=a[i]; }{void *t=a; a=b; b=t;}
	memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i]>>(3<<3))&0xFF]++; } for(int i=0xFE; i>=0; i--) c[i]+=c[i+1]; for(int i=count-1; i>=0; i--){ b[--c[(a[i]>>(3<<3))&0xFF]]=a[i]; }{void *t=a; a=b; b=t;}
	free(b);
}
int32_t nextint(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); int s=0; if(c=='-'){s=1;c=getchar_unlocked();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }

int d[100000];

int bs(int k, int l, int r){
	while(l<=r){
		int m=(l+r)/2;
		if(d[m]<k) r=m-1; else l=m+1;
	}
	return l;
}

int main(void){
	int n=nextint();
	for(int i=0; i<n; i++){
		d[i]=-1;
		int a=nextint();
		int j=bs(a,0,i-1);
		d[j]=a;
	}
	int ans=bs(0,0,n-1);
	printf("%d\n", ans);
}