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.*;
import java.math.*;

public class A {
	public static void main(String[] args) throws IOException {

		/**/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		/*/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(new FileInputStream("src/a.in"))));
		/**/
		
		int n = sc.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		int[] c1 = new int[n+2];
		int[] c2 = new int[n+2];
		for (int i = 0; i < n; i++) {
			c1[i+1] = c1[i]+2-a[i];
			c2[i+1] = c2[i]+a[i]-1;
		}
		c1[n+1] = c1[n];
		c2[n+1] = c2[n];
		int[] beg = new int[n+1];
		int[] end = new int[n+1];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j <= i; j++) {
				beg[i] = Math.max(beg[i], c1[j+1]+c2[i+1]-c2[j]);
			}
			for (int j = i; j <= n; j++) {
				end[i] = Math.max(end[i], c1[j+1]-c1[i]+c2[n]-c2[j]);
			}
		}
		beg[n] = beg[n-1];
		int max = 0;
		for (int i = 0; i < n; i++) {
			max = Math.max(max, beg[i]+end[i+1]);
		}
		System.out.println(max);
	}
}