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 Kotlin (1.3.71) (judge verdict: Runtime Error)

fun main() {
    val n= readLine()!!.toInt()
    val a= IntArray(n){ readLine()!!.toInt()}
    var dp= mutableListOf(IntArray(n){ Int.MAX_VALUE})
    var tmp= mutableListOf(-1)
    for (i in a){
        for (j in dp.indices){
            val ind = dp[j].binarySearch(i)
            //                新しく挿入
//            長さ更新
            if (tmp[j]<-ind-1){
                tmp[j]=-ind-1
                dp[j][-ind-1]=i
                break
            }
//                色の追加
            if ( j==dp.size-1){
                dp.add(dp[j].clone())
                if (ind>-1){
                    tmp.add(-ind-1)
                }else{
                    dp[j+1][-ind-1]=i
                    tmp.add(-ind-1)
                }
                break
            }
        }

//        if (-ind-1<tmp){cnt++}
    }
//    dp.forEach{ println(it.toList())}
    println(dp.size)
}