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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}

/*----------------------priorityqueue-----------------------------------------------------------*/
// kind = max?1:0
ll HeapValue(ll a){
    return a?-INF:INF;
}
ll Heapcmp(ll a, ll b, ll kind){
    return kind ? (a<b?1:0) : (a>b?1:0);
}
ll heap[MAX_N];
void pushHeap(ll val, ll len , ll kind){
    ll i=len; heap[i]=val;
    while(i!=0){
        if( Heapcmp( heap[i], heap[(i-1)/2] , ABS(kind-1) )  ) swap(&heap[i],&heap[(i-1)/2]);
        i--;i/=2;
    }
}
void popHeap(ll *val, ll len, ll kind){
    ll i=0; *val=heap[0]; heap[0]=HeapValue(kind);
    swap(&heap[0],&heap[len-1]);
    while(len>i*2+1){
        if(i*2+2<len && Heapcmp(heap[i],heap[i*2+2],kind) && Heapcmp(heap[i*2+1],heap[i*2+2],kind) ){
            swap(&heap[i],&heap[i*2+2]);
            i*=2;i+=2;
        }else{
            if( Heapcmp(heap[i],heap[i*2+1],kind) ){
                swap(&heap[i],&heap[i*2+1]);
                i*=2;i++;
            }else break;
        }
    }
}
/*---------------------------------------------------------------------------------*/

int main(void){
    // char ;
    ll n;
    scanf("%lld",&n);
    ll a[n],temp[n];
    rep(i,0,n){
        scanf("%lld",&a[i]);
    }
    ll tempsize,valmin=a[0];
    ll size=0;
    pushHeap(a[0],size,1);
    size++;
    rep(i,1,n){
        if(valmin>=a[i]){
            pushHeap(a[i],size,1);
            size++;
            valmin=a[i];
        }else{
            tempsize=0;
            ll tmp;
            while(heap[0]>=a[i]){
                popHeap(&tmp,size,1);
                temp[tempsize]=tmp;
                tempsize++;
                size--;
            }
            popHeap(&tmp,size,1);
            size--;
            pushHeap(a[i],size,1);
            size++;
            rep(j,0,tempsize){
                pushHeap(temp[j],size,1);
                size++;
            }
        }
        // printf("%lld ",size);
    }

    
    printf("%lld\n",size);
    return 0;
}