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>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define max(a, b)	(((a) > (b)) ? (a) : (b))		/* ２個の値の最大値 */
#define min(a, b)	(((a) < (b)) ? (a) : (b))		/* ２個の値の最小値 */
#define ENTER		printf("\n")					/* 改行プリント */
int DBG = 1;										/* デバッグプリント 提出時は0 */
/* main *************************************************************/
int a[100001],dp[100001];

int main()
{
	int		c,i,j,k,n,x,y,ans;
	
	scanf("%d", &n);
	for(i=1;i<=n;i++) scanf("%d", &a[i]);
	
	if (n>1) {
		ans = dp[1] = 1;
		for(i=2;i<=n;i++) {
			if (a[i-1] < a[i]) {
				dp[i] = dp[i-1];
			} else {
				for(j=i-1;j>=1;j--) {
					if (dp[j]!=dp[i-1]) break;
				}
				x = 1;
				for(;j>=1;j--) {
					if (a[j]<a[i]) {
						dp[i] = dp[j];
						x = 0;
						break;
					}
				}
				if (x) dp[i] = ++ans;
			}
		}
	}

	printf("%d\n",ans);


	return 0;
}