codeforce 933A Twisty Movement(dp+LIS)

A. A Twisty Movement

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar 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, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.

Input

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

The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 ≤ ai ≤ 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

Copy

4
1 2 1 2

output

4

input

Copy

10
1 1 2 2 2 1 1 2 2 1

output

9

Note

In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.

In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.

題意：給出n個數（均爲1或2），求反轉一個區間後，最大可以得到最長非減子序列的長度

要求最長非減子序列，首先想到LIS,可以正向和反向分別預處理出每一個區間的最長非減子序列長度，設正向爲dp1[i][j],反向爲dp2[i][j],那麼所求即爲max(dp1[1][n]-dp1[i][j]+dp2[i][j]),枚舉i和j即可

#include<bits/stdc++.h>
using namespace std;
int a[2005];
int dp1[2005][2005],dp2[2005][2005],tmp[2005];
int main()
{
    int n,len,maxx=0;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    for(int i=1; i<=n; i++)
    {
        len=0;
        memset(tmp,0,sizeof(tmp));
        for(int j=i; j<=n; j++)
        {
            if(a[j]>=tmp[len])
                tmp[++len]=a[j];
            else
            {
                int pos=upper_bound(tmp+1,tmp+len+1,a[j])-tmp;
                tmp[pos]=a[j];
            }
            dp1[i][j]=len;
        }
    }
    for(int i=n; i>=1; i--)
    {
        len=0;
        memset(tmp,0,sizeof(tmp));
        for(int j=i; j>=1; j--)
        {
            if(a[j]>=tmp[len])
                tmp[++len]=a[j];
            else
            {
                int pos=upper_bound(tmp+1,tmp+len+1,a[j])-tmp;
                tmp[pos]=a[j];
            }
            dp2[j][i]=len;
        }
    }
    for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)
        maxx=max(maxx,dp1[1][n]-dp1[i][j]+dp2[i][j]);
    cout<<maxx<<endl;
    return 0;
}