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 MOD 1000000007
//#define MOD 998244353
#define INF 1145141919810893364
typedef pair<int,int> PP;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
#define int ll
#define setdouble setprecision
#define REP(i,n) for(int i=0;i<(n);++i)
#define OREP(i,n) for(int i=1;i<=(n);++i)
#define RREP(i,n) for(int i=(n)-1;i>=0;--i)
#define GOODBYE cout << -1 << endl;return 0
#define MM <<" "<<
#define Endl endl


signed main(void){
    int N;
    int A[114514];
    int B[114514]={};
    int C[114514]={};
    cin >> N;
    REP(i,N){
        cin >> A[i];
    }
    C[N-1]=A[N-1];
    RREP(i,N-1){
        C[i]=max(C[i+1],A[i]);
    }
    int Ans=0;
    int r=0;
    while(r<N){
        Ans++;
        int now=-1;
        for(int i=0;i<N;i++){
            if(B[i]==0 && now<A[i]){
                now=A[i];
                B[i]=1;
                r++;
            }
            if(C[i]<now)break;
        }
        C[N]=-1;
        RREP(i,N-1){
            if(B[i]==0){
                C[i]=max(C[i+1],A[i]);
            }else{
                C[i]=C[i+1];
            }
        }
    }
    cout << Ans << endl;
    return 0;
}