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.io.*;

public class p934C 
{
	static int[] a;
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer st;
		int size = Integer.parseInt(f.readLine());
		a = new int[size];
		st = new StringTokenizer(f.readLine());
		for(int i = 0; i < size; i++)
			a[i] = Integer.parseInt(st.nextToken());
		int result = count(0, size - 1);
		for(int i = 0; i < size - 1; i++)
			result = Math.max(result, count(0, i) + count(i + 1, size - 1));
		out.println(result);
		out.close();
	}
	
	public static int count(int s, int e)
	{
		int right, left, result;
		left = right = result = 0;
		for(int i = s; i <= e; i++)
			if(a[i] == 2)
				right++;
		result = right;
		for(int i = s; i <= e; i++)
		{
			if(a[i] == 2)
				right--;
			else
				left++;
			result = Math.max(result, left + right);
		}
		return result;
	}
}