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.
//package albert.codeforces.atwistymovement933a;

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = scanner.nextInt();
        }

        //dp[0] = 1
        //dp[1] = 12
        //dp[2] = 121
        //dp[3] = 1212
        int[] dp = new int[4];
        for (int i = 0; i < n; i++) {
            int v = array[i];
            if (v == 1) {
                dp[0]++;
                dp[2] = Math.max(dp[2], dp[1]) + 1;
            } else {
                dp[1] = Math.max(dp[0], dp[1]) + 1;
                if (dp[2] > 0) {
                    dp[3] = Math.max(dp[2], dp[3]) + 1;
                }
            }
        }
        System.out.println(Math.max(Math.max(dp[0], dp[1]), Math.max(dp[3], dp[2])));
    }

}
