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.*;
import java.lang.*;
import java.math.*;
public class A {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        // Scanner scan = new Scanner(System.in);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(bf.readLine());
        StringTokenizer st = new StringTokenizer(bf.readLine());
        int[] a = new int[n+2];
        for(int i=0; i<n; i++) a[i+1] =  Integer.parseInt(st.nextToken());
        a[0] = 1;
        a[n+1] = 2;
        int[][] dp = new int[n+2][4];
        if(a[0] == 1) {
          dp[0][0] = 1;
        }
        for(int i=1; i<n+2; i++) {
          dp[i][0] = dp[i-1][0];
          dp[i][1] = dp[i-1][1];
          dp[i][2] = dp[i-1][2];
          dp[i][3] = dp[i-1][3];
          if(a[i] == 1) {
            dp[i][0]++;
            dp[i][2]++;
            dp[i][2] = Math.max(dp[i][2], dp[i][1]+1);
          }
          else {
            dp[i][1]++;
            dp[i][3]++;
            dp[i][1] = Math.max(dp[i][1], dp[i][0]+1);
            dp[i][3] = Math.max(dp[i][3], dp[i][2]+1);
          }
        }
        out.println((Math.max(Math.max(dp[n+1][0], dp[n+1][1]), Math.max(dp[n+1][2], dp[n+1][3]))-2));
        out.close(); System.exit(0);
    }


}
