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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a),i##formax=(b);i<i##formax;i++)
#define FORR(i,a,b) for(int i=(a),i##formax=(b);i>=i##formax;i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pcnt __builtin_popcount
#define sz(x) (int)(x).size()
#define maxs(x,y) x=max(x,y)
#define mins(x,y) x=min(x,y)
#define show(x) cout<<#x<<" = "<<x<<endl;
#define all(a) (a.begin()),(a.end())
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define perm(c) sort(all(c));for(bool c##p=1;c##p;c##p=next_permutation(all(c)))
#define bitComb(a,n,k) for(int a##x,a##y,a=(1<<k)-1;a<(1<<n);a##x=a&-a,a##y=a+a##x,a=(((a&~a##y)/a##x)>>1)|a##y)
#define uniq(v) sort(all(v));v.erase(unique(all(v)), v.end())
#define bit(n) (1LL<<(n))
#define randInt(l,r) (uniform_int_distribution<ll>(l,r)(rnd))
#define randDouble(l,r) (uniform_real_distribution<double>(l,r)(rnd))
#define dout(d) printf("%.12f\n",d)

typedef long long ll;
typedef __int128_t lll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<class T>using V=vector<T>;
template<class T>using VV=V<V<T>>;
template<class T,class Y>ostream& operator<<(ostream& o,const pair<T,Y>& p){return o<<"("<<p.fi<<", "<<p.se<<")";}
template<typename A,size_t N,typename T>void Fill(A (&array)[N],const T&v){fill((T*)array,(T*)(array+N),v);}
lll gcd(lll a,lll b,lll &x,lll &y){if(!b){x=1;y=0;return a;}lll d=gcd(b,a%b,y,x);y-=a/b*x;return d;}
ll modInv(ll a,ll m){lll x,y;gcd(a,m,x,y);return (x%m+m)%m;}
ll modPow(lll a,lll n,ll m){if(!a)return a;lll p=1;for(;n>0;n>>=1,a=a*a%m)if(n&1)p=p*a%m;return(ll)p;}
bool isPrime(ll n){if(n<2||n%2==0)return n==2;lll t=n-1,d=t/(t&-t);for(lll a:{2,325,9375,28178,450775,9780504,1795265022})if(a%n){for(t=d,a=modPow(a,t,n);t!=n-1&&a!=1&&a!=n-1;a=a*a%n,t=t*2%n);if(a!=n-1&&t%2==0)return 0;}return 1;}
//V<ll>fac,rfac;void setFac(int n,int mod){fac=rfac=V<ll>(++n);fac[0]=1;FOR(i,1,n)fac[i]=fac[i-1]*i%mod;rfac[n-1]=modInv(fac[n-1],mod);FORR(i,n-1,1)rfac[i-1]=rfac[i]*i%mod;}
const int IINF = 1e9+6;
const ll LINF = 1e18;
const int MOD = 1e9+7;
const double PI = acos(-1);
const double EPS = 1e-10;
static random_device rd;
static mt19937 rnd(rd());

const int N = 1e5;
int n;
multiset<int> s;
main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ans = 0;
  cin >> n;
  FOR(i, 0, n){
    int a;
    cin >> a;
    s.insert(a);
    auto itr = s.find(a);
    if(itr == s.begin()){
      ans++;
    }else{
      s.erase(--itr);
    }
  }
  cout << ans << endl;
}