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)

/*
 * main.c
 *
 *  Created on: 2019/07/07
 *      Author: family
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
typedef unsigned long long int uint64_t;
typedef long long int int64my_t;

//qsort(a, 3, sizeof(int), sort_inc);
int sort_inc(const void *a, const void *b)
{
    return (*(int*)a - *(int*)b);
}
int sort_dec(const void* a, const void* b)
{
    return (*(int*)b - *(int*)a);
}
#define FIXVAL (1000000000 + 7)

int A[100001] = {0};

int main()
{
	int N = 0, i = 0, color = 1, same = 0;
	scanf("%d", &N);
	for (i = 0; i < N; i++) {
		scanf("%d", &A[i]);
	}
	for (i = 0; i < N; i++) {
		if (i >= 1) {
			if (A[i-1] < A[i]) {

			} else if (A[i-1] == A[i]) {
				color++;
			} else {
				same = 1;
			}
		}
	}
	printf("%d\n", color + same);
    return 0;
}