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.
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */

public class Main{
	public static void main(String[] args){
		Scanner scan=new Scanner(System.in);
		Sequence seq=new Sequence();
		seq.n=scan.nextInt();
		seq.A=new int[seq.n];
		for(int i=0;i<seq.n;i++){
			seq.A[i]=scan.nextInt();		
		}
		System.out.println(seq.doit());
	}
}

class Sequence{
	int n;
	int[] A;
	int doit(){
		int[] none=new int[n];
		int[] ntwo=new int[n];
		none[0]=((A[0]==1)?1:0);
		ntwo[0]=1-none[0];
		for(int i=1;i<n;i++){
			none[i]=none[i-1]+((A[i]==1)?1:0);
			ntwo[i]=i+1-none[i];
		}
		int res=0;
		for(int i=0;i<n;i++){
			int tot=none[i]+(ntwo[n-1]-ntwo[i]);
			if(tot>res){res=tot;}
		}
		if(ntwo[n-1]>res){res=ntwo[n-1];}
		int[][] dp=new int[n][n];
		for(int l=0;l<n;l++){
			dp[l][l]=1;
			for(int r=l+1;r<n;r++){
				if(A[r]==1){dp[l][r]=dp[l][r-1]+1;}
				else{
					int amt=((l==0)?ntwo[r]:(ntwo[r]-ntwo[l-1]));
					dp[l][r]=(int)Math.max(dp[l][r-1],amt);
				}
			}
		}
		for(int l=0;l<n;l++){
			for(int r=l;r<n;r++){
				int x=((l==0)?0:none[l-1]);
				int y=ntwo[n-1]-ntwo[r];
				res=(int)Math.max(res,x+y+dp[l][r]);
			}
		}
		return res;
	}
}

