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.Scanner;

public class A {
    public static void main(String... args) {
        final Scanner sc = new Scanner(System.in);
        final int n = sc.nextInt();
        final int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = sc.nextInt();
        final int[] ones = new int[n + 1];
        final int[] twos = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (a[i] == 1) {
                ones[i + 1] = ones[i] + 1;
                twos[i + 1] = twos[i];
            } else {
                ones[i + 1] = ones[i];
                twos[i + 1] = twos[i] + 1;
            }
        }
        final int[] left = new int[n + 1];
        for(int i = 0; i <= n; i++)
            for(int j = 0; j <= i; j++)
                left[i] = Math.max(left[i], ones[j] + twos[i] - twos[j]);
        final int[] right = new int[n + 1];
        for(int i = 0; i <= n; i++)
            for(int j = i; j <= n; j++)
                right[i] = Math.max(right[i], ones[j] - ones[i] + twos[n] - twos[j]);
        int max = 0;
        for(int i = 0; i <= n; i++)
            max = Math.max(max, left[i] + right[i]);
        System.out.println(max);
    }
}
