Description

A dragon symbolizes wisdom, power and wealth. On Lunar New Year’s Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.

A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence

a1,a2,...,an$a_1,a_2,...,a_n$ .

Little Tommy is among them. He would like to choose an interval [l,r](1≤l≤r≤n)$[l,r](1\le l\le r\le n)$, then reverse al,al+1,...,ar$a_l,a_{l+1},...,a_r$ so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

A non-decreasing subsequence is a sequence of indices

p1,p2,...,pk$p_1,p_2,...,p_k$ , such that p1<p2<...<pk$p_1<p_2<...<p_k$ and ap1≤ap2≤...≤apk$a_{p_1}\le a_{p_2}\le ...\le a_{p_k}$ . The length of the subsequence is k$k$ .

Input

The first line contains an integer n(1≤n≤2000)$n(1\le n\le 2000)$ , denoting the length of the original sequence.

The second line contains

n$n$ space-separated integers, describing the original sequence a1,a2,...,an(1≤ai≤2,i=1,2,...,n)$a1,a2,...,an(1\le ai\le 2,i=1,2,...,n)$ .

Output

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

Examples input

4
1 2 1 2

Examples output

4

題意

在一個只包含 1,2$1,2$ 的序列中，翻轉其中任意一個區間，求此時最大的 LIS 。

思路

正著倒著預處理出每一段的 LIS，分別記為 L[i][j],R[i][j]$L[i][j],R[i][j]$

然後開始列舉，翻轉一個區間相當於減去這段區間的貢獻，然後加上其翻轉以後的

此時 LIS 為 L[0][n−1]−L[i][j]+R[i][j]$L[0][n-1]-L[i][j]+R[i][j]$

AC 程式碼

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
typedef __int64 LL;
const int maxn = 2e3+10;

int n;
int a[maxn];
int L[maxn][maxn],R[maxn][maxn];

int main()
{
    IO;
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>a[i];
    for(int i=0; i<n; i++)
    {
        int dp1 = 0, dp2 = 0;
        for(int j=i; j<n; j++)
        {
            if(a[j]==1)
                ++dp1;
            else
                dp2 = max(dp1,dp2) + 1;
            L[i][j] = max(dp1,dp2);
        }
        dp1 = dp2 = 0;
        for(int j=i; j<n; j++)
        {
            if(a[j]==2)
                ++dp1;
            else
                dp2 = max(dp1,dp2) + 1;
            R[i][j] = max(dp1,dp2);
        }
    }
    int ans = 0;
    for(int i=0; i<n; i++)
        for(int j=i; j<n; j++)
            ans = max(ans,L[0][n-1] - L[i][j] + R[i][j]);
    cout<<ans<<endl;
    return 0;
}