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 aTwistyMovement {
	static int n;
	static int[] arr, num1s;
	public static void main(String[] args) {
		FS sc = new FS();
		n = sc.nextInt();
		arr = new int[n];
		for(int i = 0; i < n; ++i) arr[i] = sc.nextInt() - 1;
		
		num1s = new int[n + 1];
		int l0 = 0, l1 = 0;
		for(int i = n - 1; i >= 0; --i) {
			if(arr[i] == 0) l0 = Math.max(l0, l1) + 1;
			else ++l1;
			
			num1s[i] = l1;
		}
		
		int out = Math.max(l0, l1);
		
		// brute force flipping range [i, i + 1]
		int num0s = 0;
		for(int l = 0; l < n; ++l) {
			l0 = l1 = 0;
			for(int r = l; r < n; ++r) {
				if(arr[r] == 0) l0 = Math.max(l0, l1) + 1;
				else ++l1;
				
				int curr = num0s + Math.max(l0, l1) + num1s[r + 1];
				out = Math.max(out, curr);
			}
			
			if(arr[l] == 0) ++num0s;
		}
		
		System.out.println(out);
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}