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 Main {

    public static void main(String[] args) throws Exception {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        int[] a=new int[n];
        for (int i=0;i<n;i++) a[i]=scanner.nextInt();

        int[][][] number=new int[n][n][2]; // 0: at least one 1; 1: all 2

        int[] before=new int[n], after=new int[n];
        for (int i=0;i<n;i++) {
            before[i]=2-a[i]+(i>0?before[i-1]:0);
        }
        for (int i=n-1;i>=0;i--) {
            after[i]=a[i]-1+(i<n-1?after[i+1]:0);
        }

        int max=0;
        for (int i=0;i<n;i++)
            max=Math.max(max, before[i]+after[i]);

        for (int i=0;i<n;i++) {
            for (int j=i;j<n;j++) {
                if (i==j) {
                    number[i][j][a[i]-1]=1;
                }
                else {
                    if (a[j]==1) {
                        number[i][j][0]=Math.max(number[i][j-1][0], number[i][j-1][1])+1;
                        number[i][j][1]=number[i][j-1][1];
                    }
                    else {
                        number[i][j][0]=number[i][j-1][0];
                        number[i][j][1]=number[i][j-1][1]+1;
                    }
                }

                max=Math.max(max, (i==0?0:before[i-1]) + Math.max(number[i][j][0], number[i][j][1]) + (j==n-1?0:after[j+1]));
            }
        }

        System.out.println(max);

    }


}