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.*;
public class Ishu
{
    static Scanner scan = new Scanner(System.in);
    static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out)); 
    static void tc() throws Exception
    {
    int n = scan.nextInt();
    int[] a = new int[n + 1];
    int i;
    for(i=1;i<=n;++i)
        a[i] = scan.nextInt();
        
    int[] one = new int[n + 2];
    int[] two = new int[n + 2]; 
    int[][] pre = new int[2][n + 2];
    int[][] suf = new int[2][n + 2];
    int[][] prede = new int[2][n + 2];
    
    for(i=1;i<=n;++i)
        {
        if(a[i] == 1)
            {
            one[i] = one[i-1] + 1;
            two[i] = two[i-1];
            }
        else
            {
            one[i] = one[i-1];
            two[i] = two[i-1] + 1;
            }
        }
    for(i=1;i<=n;++i)    
        {
        if(a[i] == 1)
            {
            pre[0][i] = pre[0][i-1] + 1;
            pre[1][i] = pre[1][i-1];
            }
        else
            {
            pre[0][i] = pre[0][i-1]; 
            pre[1][i] = Math.max(pre[0][i-1], pre[1][i-1]) + 1;
            }
        }
        
    for(i=n;i>=1;--i)   
        {
        if(a[i] == 1)
            {
            prede[0][i] = Math.max(prede[0][i+1], prede[1][i+1]) + 1;
            prede[1][i] = prede[1][i+1];
            }
        else
            {
            prede[0][i] = prede[0][i+1];
            prede[1][i] = prede[1][i+1] + 1;
            }
        }
        
    int ans = 0;
    int T = 0;
    int O = 0;
    for(int l=1;l<=n;++l)    
        {
        T = 0;
        O = 0;
        for(int r=l;r<=n;++r)
            {
            if(a[r] == 1)
                O = Math.max(T, O) + 1;
            else
                ++T;
                
            int cr = 0;
            cr += pre[0][l-1];
            cr += one[r] - one[l-1];
            cr += Math.max(prede[0][r+1], prede[1][r+1]);
            ans = Math.max(ans, cr);
            
            cr = 0;
            cr += pre[0][l-1];
            cr += Math.max(T, O);
            cr += prede[1][r+1];
            ans = Math.max(ans, cr);
            
            cr = 0;
            cr += Math.max(pre[0][l-1], pre[1][l-1]);
            cr += two[r] - two[l-1];
            cr += prede[1][r+1];
            ans = Math.max(ans, cr);
            }
        }
    output.write(ans + "\n");
    output.flush();
    }
    public static void main(String[] args) throws Exception
    {
    int t = 1;
    //t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}
