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].
題目連結
參考題解
氣泡排序，每一次交換的時候，這兩個數字節點之間就連一條邊，到最後有幾個沒有連線的節點。這樣的話如果是順序正確那麼就不會連邊，所以直接求最長非降子序列就可以了。

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
using namespace std;
int a[100005];
int b[100005];
int n;

int findd(int len,int p)   //二分查詢<=p的位置
{
    int l,r,mid;
    l=1,r=len,mid=(l+r)>>1;
    while(l<=r)
    {
        if(p>b[mid]) l=mid+1;
        else if(p<b[mid]) r=mid-1;
        else return mid;
        mid=(l+r)>>1;
    }
    return l;
}

int LIS()
{
    int i,j,len=1;
    b[1]=a[0];
    for(i=1;i<n;i++)
    {
        j=findd(len,a[i]);
        b[j]=a[i];  //b[j]是指長度為j最大元素的值
        if(j>=len) len=j;
    }
    return len;
}

int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        printf("%d\n",LIS());
    }
    return 0;
}