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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define uli unsigned long long int
#define INF 999999999999999999
#define rep(i,m,n) for(lli i = m;i < n;i++)
#define rrep(i,m,n) for(lli i=m-1;i>=n;i--)
#define pb(n) push_back(n)
#define UE(N) N.erase(unique(N.begin(),N.end()),N.end());
#define Sort(n) sort(n.begin(), n.end())
#define Rev(n) reverse(n.begin(),n.end())
#define Out(S) cout << S << endl
#define NeOut(S) cout << S
#define HpOut(S) cout << setprecision(30) << S << endl
#define Vec(K,L,N,S) vector<L> K(N,S)
#define DV(K,L,N,M,S) vector<vector<L>> K(N,vector<L>(M,S))
#define TV(K,L,N,M,R,S) vector<vector<vector<L>>> K(N,vector<vector<L>>(M,vector<L>(R,S)))
#define pint pair<lli,lli>
#define paf(L,R) pair<L,R>
#define mod 1000000007
#define MAX 5100000
#define ALL(a)  a.begin(),a.end()
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
int main(){
  lli A,B,C,D,E,F,N,M,K,L,R,X,Y,Z,H,W,sum=0,num=0,flag=0;string S,T;
  cin >> N;
  Vec(P,lli,N,0);
  rep(i,0,N)cin >> P[i];
  Vec(dist,lli,N,-1);
  queue<lli> que;
  dist[0]=1;
  que.push(P[0]);
  sum=1;
  while(!que.empty()){
    flag=0;
    auto Xq=que.front();que.pop();
    rep(i,0,N){
      if(dist[i]!=-1)continue;
      if(P[i]<=Xq)continue;
      dist[i]=sum;
      que.push(P[i]);
      flag=1;
      break;
    }
    if(!flag){
      rep(i,0,N)if(dist[i]==-1){
        dist[i]=sum+1;
        sum++;
        que.push(P[i]);
        break;
      }
    }
  }
  Sort(dist);
  Out(dist[N-1]);
}