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

4
1 2 1 2


Output

4


Input

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.
import java.io.*;
import java.util.*;
public class A933 
{
    public static void main(String args[])throws IOException
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int arr[]=new int[n+1];
        for(int i=1;i<=n;i++)
        {
            arr[i]=sc.nextInt();
        }
        int dp1[]=new int[n+1];
        int dp2[]=new int[n+1];
        int dec[]=new int[n+1];
        int dep[]=new int[n+1];
        int max1[]=new int[n+1];
        int max2[]=new int[n+2];
        for(int i=1;i<=n;i++)
        {
            dp1[i]=1;
            dep[i]=1;
            for(int j=i-1;j>=1;j--)
            {
                if(arr[i]>=arr[j])
                {
                    dp1[i]=(int)Math.max(dp1[i], dp1[j]+1);
                    max1[i]=(int)Math.max(max1[i-1],dp1[i]);
                }
                if(arr[i]<=arr[j] && dec[j]+1 > dec[i])
                {
                    dec[i]=dec[j]+1;
                    dep[i]=j;
                }
            }
        }
        for(int i=n;i>=1;i--)
        {
            dp2[i]=1;
            for(int j=i+1;j<=n;j++)
            {
                if(arr[i]<=arr[j])
                {
                    dp2[i]=(int)Math.max(dp2[i],dp2[j]+1);
                    max2[i]=(int)Math.max(max2[i+1],dp2[i]);
                }
            }
        }
        int max=0;
        
        for(int i=1;i<=n;i++)
        {
            int ind=i;
            int c=0;
            while(ind!=dep[ind])
            {
                c++;
                max=(int)Math.max(max,c+max1[ind-1]+max2[i+1]);
                ind=dep[ind];
            }
            c++;
            max=(int)Math.max(max,c+max1[ind-1]+max2[i+1]);
        }
        System.out.println(max);
        
    }
}
