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.*;

public class Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int n = sc.nextInt();
              int arr[] = new int[n];
              for(int i = 0 ; i < n ; i++)
              {
                     arr[i] = sc.nextInt();
              }
              
              int pre[][] = new int[n][2];
              if(arr[0] == 1)
              {
                     pre[0][0] = 1;
                     pre[0][1] = 0;
              }
              
              else
              {
                     pre[0][0] = 0;
                     pre[0][1] = 1;
              }
              for(int i = 1 ; i < n ; i++)
              {
                     if(arr[i] == 1)
                     {
                            pre[i][0] = pre[i-1][0]+1;
                            pre[i][1] = pre[i-1][1];
                     }
                     
                     else
                     {
                            pre[i][0] = pre[i-1][0];
                            pre[i][1] = Math.max(pre[i-1][0] , pre[i-1][1])+1;
                     }
              }
              
              int post[][] = new int[n][2];
              if(arr[n-1] == 1)
              {
                     post[n-1][0] = 1;
                     post[n-1][1] = 0;
              }
              
              else
              {
                  post[n-1][0] = 0;
                     post[n-1][1] = 1;    
              }
              
              for(int i = n-2 ; i >= 0 ; i--)
              {
                     if(arr[i] == 1)
                     {
                            post[i][1] = post[i+1][1];
                            post[i][0] = Math.max(post[i+1][0] , post[i+1][1])+1;
                     }
                     
                     else
                     {
                           post[i][1] = post[i+1][1]+1;
                           post[i][0] = post[i+1][0];
                     }
              }
              int max = 0;
              for(int i = 0 ; i < n ; i++)
              {
          int sm = Math.max(pre[i][0] , pre[i][1]);
          if(i+1 < n )
          {
                 sm += Math.max(post[i+1][0] , post[i+1][1]);
          }
          max = Math.max(max , sm);
              }
              System.out.println(max);
              }
}