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.util.*;
import java.time.*;
import java.io.*;
import java.lang.Math;
public class test {
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int[] a=new int[2001],cnt1=new int[2001],cnt2=new int[2001];
		int[][] f=new int[2001][2001];
		
		for(int i=1;i<=n;++i)
			a[i]=in.nextInt();
		
		cnt1[0]=0;
		cnt2[0]=0;
		for(int i=1;i<=n;++i)
		{
			if(a[i]==1)
			{
				cnt1[i]=cnt1[i-1]+1;
				cnt2[i]=cnt2[i-1];
			}
			else
			{
				cnt1[i]=cnt1[i-1];
				cnt2[i]=cnt2[i-1]+1;
			}
		}
		
		for(int i=1;i<=n;++i)
			for(int j=i;j<=n;++j)
			{
				if(i==j)
					f[i][j]=0;
				else if(a[j]==1)
				{
					f[i][j]=Math.max(f[i][j-1],cnt2[j-1]-cnt2[i-1])+1;
					if(f[i][j]==1)
						f[i][j]=0;
				}
				else
					f[i][j]=f[i][j-1];
			}
		
		int maxV=Math.max(cnt1[n],cnt2[n]);
		for(int i=1;i<=n;++i)
			for(int j=i;j<=n;++j)
			{
				if(i==j)
					maxV=Math.max(maxV,cnt1[i-1]+cnt2[n]-cnt2[i-1]);
				else
				{
					maxV=Math.max(maxV,cnt1[i-1]+cnt2[n]-cnt2[j]+f[i][j]);
				}
			}
		
		System.out.println(maxV);
	}
}