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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

#define REP(i,num) for(int i=0;i<(num);++i)
#define ALL(c) c.begin(),c.end()
#define PRINTALL(c) for(auto& x:c){cout<<x<<' ';}cout<<endl;

using namespace std;
using ll = long long;

template<typename T=int>
T in(){T x; cin >> x; return (x);}

vector<ll> enum_div(ll n){
	vector<ll> ret;
	for(ll i=1;i*i<=n;++i){
		if(n%i==0){
			ret.push_back(i);
			if(i*i!=n){
				ret.push_back(n/i);
			}
		}
	}
	return ret;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N=in();
	vector<pair<int,int>> p(1,make_pair(-1,-1));
	REP(i,N){
		int a = in();
		if(p.back().first==-1){
			p.back().first = a;
			p.back().second = a;
		}
		else if(p.back().second<a){
			p.back().second = a;
		}
		else{
			p.emplace_back(make_pair(-1,-1));
			p.back().first = a;
			p.back().second = a;
		}
	}
	
	//sort(ALL(p),[](pair<int,int>& lhs,pair<int,int>& rhs){return lhs.first<rhs.first;});
	auto comp = [](pair<int,int>& lhs,pair<int,int>& rhs){return lhs.second>rhs.second;};
	priority_queue<pair<int,int>,vector<pair<int,int>>,decltype(comp)> pq(comp);
	pq.push(p[0]);
	int count = 1;
	for(int i=1;i<p.size();i++){
		auto q = pq.top();
		pq.pop();
		if(q.second<p[i].first){
			q.second = p[i].first;
			pq.push(q);
		}
		else{
			pq.push(q);
			pq.push(p[i]);
			count++;
		}
	}
	cout << count << endl;
	return 0;
}