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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    mod=10**9+7
    inf=10**10


    ##############################
    from bisect import bisect_left, bisect_right, insort_right
    class SquareSkipList:
        # SkipList の層数を 2 にした感じの何か
        # std::multiset の代用になる
        
        def __init__(self, values=None, sorted_=False, square=1000, seed=42):
            # values: 初期値のリスト
            # sorted_: 初期値がソート済みであるか
            # square: 最大データ数の平方根
            # seed: 乱数のシード
            ### pop関数は　for を回すので重め、使うなら square パラメータを大きめにするべき
            
            inf = float("inf")
            self.square = square
            if values is None:
                self.rand_y = seed
                self.layer1 = [inf]
                self.layer0 = [[]]
            else:
                self.layer1 = layer1 = []
                self.layer0 = layer0 = []
                if not sorted_:
                    values.sort()
                y = seed
                l0 = []
                for v in values:
                    y ^= (y & 0x7ffff) << 13
                    y ^= y >> 17
                    y ^= (y & 0x7ffffff) << 5
                    if y % square == 0:
                        layer0.append(l0)
                        l0 = []
                        layer1.append(v)
                    else:
                        l0.append(v)
                layer1.append(inf)
                layer0.append(l0)
                self.rand_y = y

        def add(self, x):  # 要素の追加  # O(sqrt(n))
            # xorshift
            y = self.rand_y
            y ^= (y & 0x7ffff) << 13
            y ^= y >> 17
            y ^= (y & 0x7ffffff) << 5
            self.rand_y = y

            if y % self.square == 0:
                layer1, layer0 = self.layer1, self.layer0
                idx1 = bisect_right(layer1, x)
                layer1.insert(idx1, x)
                layer0_idx1 = layer0[idx1]
                idx0 = bisect_right(layer0_idx1, x)
                layer0.insert(idx1+1, layer0_idx1[idx0:])  # layer0 は dict で管理した方が良いかもしれない  # dict 微妙だった
                del layer0_idx1[idx0:]
            else:
                idx1 = bisect_right(self.layer1, x)
                insort_right(self.layer0[idx1], x)

        def remove(self, x):  # 要素の削除  # O(sqrt(n))
            # x が存在しない場合、x 以上の最小の要素が削除される
            idx1 = bisect_left(self.layer1, x)
            layer0_idx1 = self.layer0[idx1]
            idx0 = bisect_left(layer0_idx1, x)
            if idx0 == len(layer0_idx1):
                del self.layer1[idx1]
                self.layer0[idx1] += self.layer0.pop(idx1+1)
            else:
                del layer0_idx1[idx0]

        def search_higher_equal(self, x):  # x 以上の最小の値を返す  O(log(n))
            idx1 = bisect_left(self.layer1, x)
            layer0_idx1 = self.layer0[idx1]
            idx0 = bisect_left(layer0_idx1, x)
            if idx0 == len(layer0_idx1):
                return self.layer1[idx1]
            return layer0_idx1[idx0]

        def search_higher(self, x):  # x を超える最小の値を返す  O(log(n))
            idx1 = bisect_right(self.layer1, x)
            layer0_idx1 = self.layer0[idx1]
            idx0 = bisect_right(layer0_idx1, x)
            if idx0 == len(layer0_idx1):
                return self.layer1[idx1]
            return layer0_idx1[idx0]

        def search_lower(self, x):  # x 未満の最大の値を返す  O(log(n))
            #x未満のものがない場合，infが返るかも！！！  
            idx1 = bisect_left(self.layer1, x)
            layer0_idx1 = self.layer0[idx1]
            idx0 = bisect_left(layer0_idx1, x)
            if idx0 == 0:  # layer0_idx1 が空の場合とすべて x 以上の場合
                return self.layer1[idx1-1]
            return layer0_idx1[idx0-1]

        def pop(self, idx):
            # 小さい方から idx 番目の要素を削除してその要素を返す（0-indexed）
            # O(sqrt(n))
            # for を回すので重め、使うなら square パラメータを大きめにするべき
            layer0 = self.layer0
            s = -1
            for i, l0 in enumerate(layer0):
                s += len(l0) + 1
                if s >= idx:
                    break
            if s==idx:
                layer0[i] += layer0.pop(i+1)
                return self.layer1.pop(i)
            else:
                return layer0[i].pop(idx-s)

        def print(self):
            print(self.layer1)
            print(self.layer0)
            
        def len(self):
            cnt=0
            for l in self.layer0:
                cnt+=len(l)
            cnt+=len(self.layer1)-1
            return cnt

    ##############################



        

            
    N=I()
    A=[0]*N
    
    for i in range(N):
        A[i]=I()
        
    ssl = SquareSkipList(square=2000)
    ssl.add(-1)#初期値入れとこ
    
    for i in range(N):
        a=A[i]
        m=ssl.search_lower(a)
        if m==-1:
            ssl.add(a)
        else:
            ssl.remove(m)
            ssl.add(a)
            
    print(ssl.len()-1)
    



main()