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.*;

public class E {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] ree = in.readLine().split("\\s++");
		int n = Integer.parseInt(ree[0]);
		String[] qwq = in.readLine().split("\\s++");
		char[] vals = new char[qwq.length];
		for(int i = 0; i < vals.length; i++) vals[i] = qwq[i].charAt(0);
		int[][] dp = new int[n][4];
		char[] self = {'1', '2', '1', '2'};
		char[] adv = {'-', '2', '1', '2'};
		int best = 0;
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < 4; j++) {
				if(i == 0) dp[i][j] = (self[j] == vals[i] ? 1 : 0);
				else {
					dp[i][j] = dp[i-1][j] + (self[j] == vals[i] ? 1 : 0);
					if(adv[j] == vals[i]) {
						if(dp[i-1][j-1] + 1 > dp[i][j]) dp[i][j] = dp[i-1][j-1] + 1;
					}
				}
				if (dp[i][j] > best) best = dp[i][j];
				//System.out.print(dp[i][j] + " ");
			}
			//System.out.println();
		}
		System.out.println(best);
	}
}