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 A933 {
	
	public static void main(String[] args) throws Exception {

		int n = Integer.parseInt(in.readLine());

		int[] arr = new int[n + 1];

		st = new StringTokenizer(in.readLine());
		for (int i = 1; i <= n; i++) {
			arr[i] = i();
		}

		int[][] dp = new int[n + 1][4];

		for (int i = 1; i <= n; i++) {
			dp[i][0] = dp[i - 1][0] + (arr[i] == 1 ? 1 : 0);
			dp[i][1] = dp[i - 1][1];
			dp[i][2] = dp[i - 1][2];
			dp[i][3] = dp[i - 1][3];


			if (arr[i] == 2){// && dp[i - 1][0] > 0) {
				dp[i][1] = Math.max(dp[i - 1][0], dp[i - 1][1]) + 1;
			}

			if (arr[i] == 1){// && dp[i - 1][1] > 0) {
				dp[i][2] = Math.max(dp[i - 1][1], dp[i - 1][2]) + 1;
			}

			if (arr[i] == 2){// && dp[i - 1][2] > 0) {
				dp[i][3] = Math.max(dp[i - 1][2], dp[i - 1][3]) + 1;
			}
		}

		// for(int[] arr2 : dp) {
		// 	out.println(Arrays.toString(arr2));
		// }
		
		int ans = Math.max(Math.max(dp[n][3], dp[n][2]), Math.max(dp[n][1] - 2, dp[n][0] == 0 ? n : dp[n][0]));
		out.println(ans);

		out.close();
	}

	static BufferedReader in;
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static {
		try {
			in = new BufferedReader(new FileReader("cf.in"));
		} catch (Exception e) {
			in = new BufferedReader(new InputStreamReader(System.in));
		}
	}
	static int i() {return Integer.parseInt(st.nextToken());}
	static double d() {return Double.parseDouble(st.nextToken());}
	static String s() {return st.nextToken();}
	static long l() {return Long.parseLong(st.nextToken());}
}
