Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a1, a2, ..., an in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode). 
    
    
      
    procedure bubbleSortGraph()  
        build a graph G with n vertices and 0 edges  
        repeat  
            swapped = false  
            for i = 1 to n - 1 inclusive do:  
                if a[i] > a[i + 1] then  
                    add an undirected edge in G between a[i] and a[i + 1]  
                    swap( a[i], a[i + 1] )  
                    swapped = true  
                end if  
            end for  
        until not swapped   
        /* repeat the algorithm as long as swapped value is true. */   
    end procedure  
    

For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 105). The next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n).

Output

Output a single integer — the answer to the problem. 

Examples

Input

3
3 1 2


Output

2

Note

Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2].
100
36 48 92 87 28 85 42 10 44 41 39 3 79 9 14 56 1 16 46 35 93 8 82 26 100 59 60 2 96 52 13 98 70 81 71 94 54 91 17 88 33 30 19 50 18 73 65 29 78 21 61 7 99 97 45 89 57 27 76 11 49 72 84 69 43 62 4 22 75 6 66 83 38 34 86 15 40 51 37 74 67 31 20 63 77 80 12 53 5 25 58 90 68 24 64 23 95 32 55 47
 5
4 2 1 3 5
 10
1 9 8 10 2 3 4 6 5 7
 50
12 24 42 43 36 3 40 29 7 34 10 13 28 9 35 23 25 21 19 4 20 18 11 38 41 48 6 46 33 17 31 37 2 30 32 44 45 5 47 49 16 15 50 27 26 14 39 22 1 8

16
 3
 6
 13

import java.util.*;
import java.io.*;
import java.math.*;
 
public class Main
{
 
    static class LIS { 
        // Binary search (note boundaries in the caller) 
        // A[] is ceilIndex in the caller 
        static int CeilIndex(int A[], int l, int r, int key) 
        { 
            while (r - l > 1) { 
                int m = l + (r - l) / 2; 
                if (A[m] >= key) 
                    r = m; 
                else
                    l = m; 
            } 

            return r; 
        } 

        static int LongestInc(int A[], int size) 
        { 
            // Add boundary case, when array size is one 

            int[] tailTable = new int[size]; 
            int len; // always points empty slot 

            tailTable[0] = A[0]; 
            len = 1; 
            for (int i = 1; i < size; i++) { 
                if (A[i] < tailTable[0]) 
                    // new smallest value 
                    tailTable[0] = A[i]; 

                else if (A[i] > tailTable[len - 1]) 
                    // A[i] wants to extend largest subsequence 
                    tailTable[len++] = A[i]; 

                else
                    // A[i] wants to be current end candidate of an existing 
                    // subsequence. It will replace ceil value in tailTable 
                    tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i]; 
            } 

            return len; 
        } 

    } 
/* This code is contributed by Devesh Agrawal*/

    public static void process(int test_number)throws IOException
    {
        LIS ob = new LIS();
        int n = ni(), arr[] = new int[n];
        for(int i = 0; i < n; i++)
            arr[i] = ni();
        pn(ob.LongestInc(arr, n));    
    }
 
    static final long mod = (long)1e9+7l;
    
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc = new FastReader();
 
        long s = System.currentTimeMillis();
        int t = 1;
        //t = ni();
        for(int i = 1; i <= t; i++)
            process(i);
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }

    static void trace(Object... o){ System.err.println(Arrays.deepToString(o)); };
    static void pn(Object o){ out.println(o); }
    static void p(Object o){ out.print(o); }
    static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
    static long nl()throws IOException{ return Long.parseLong(sc.next()); }
    static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
    static String nln()throws IOException{ return sc.nextLine(); }
    static long gcd(long a, long b)throws IOException{ return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{ return (b==0)?a:gcd(b,a%b); }
    static int bit(long n)throws IOException{ return (n==0)?0:(1+bit(n&(n-1))); }
    
    static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader(){ 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next(){ 
            while (st == null || !st.hasMoreElements()){ 
                try{ st = new StringTokenizer(br.readLine()); } catch (IOException  e){ e.printStackTrace(); } 
            } 
            return st.nextToken(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
}
