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 Solution{
   static int[] one;
   static int[] two;
   public static int un(int l,int r){
       return one[r+1] - one[l];
   }
   
   public static int deux(int l,int r){
       return two[r+1] - two[l];
   }
    
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        int[] a = new int[n];
        for(int i=0;i<n;i++) a[i] = sc.nextInt();
        one = new int[n+1];
        one[0] = 0;
        for(int i=1;i<=n;i++){
            one[i] = one[i-1] + a[i-1]%2;
        }
        two = new int[n+1];
        two[0] = 0;
        for(int i=1;i<=n;i++){
            two[i] = two[i-1] + a[i-1] - 1;
        }
        int max = (int) Math.max(un(0,n-1),deux(0,n-1));
        int[][] dp = new int[n][n];
        for(int i=0;i<n;i++){
            dp[i][i] = 1 + un(0,i-1) + deux(i+1,n-1);
            max = (int) Math.max(dp[i][i],max);
        }
        
        for(int i=0;i<n-1;i++){
            dp[i][i+1] = un(0,i-1) + deux(i+2,n-1);
            if(a[i]==a[i+1]) dp[i][i+1] += 2;
            else if(a[i]==2) dp[i][i+1] += 2;
            else if(a[i]==1) dp[i][i+1] += 1;
            max = (int) Math.max(dp[i][i+1],max);
        }
        int val = 0;
        
        for(int k=2;k<n;k++){
            for(int i=0;i<n-k;i++){
                dp[i][i+k] = dp[i+1][i+k-1];
                if(a[i]!=a[i+k]){
                    
                    if(a[i]==1) dp[i][i+k] -= 2;
                    else dp[i][i+k] += 2;
                    
                }
                
                val = un(0,i-1) + deux(i,n-1);
                dp[i][i+k] = (int) Math.max(val,dp[i][i+k]);
                val = un(0,i+k) + deux(i+k+1,n-1);
                dp[i][i+k] = (int) Math.max(val,dp[i][i+k]);
                
                max = (int) Math.max(dp[i][i+k],max);
            }
        }
        
        System.out.println(max);
        

    }
    
}