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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        r462a solver = new r462a();
        solver.solve(1, in, out);
        out.close();
    }

    static class r462a {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.ni();
            int[] a = in.na(n);
            int[] cnt1 = new int[n + 1];
            int[] cnt2 = new int[n + 1];
            for (int i = 0; i < n; i++) {
                cnt1[i + 1] = cnt1[i];
                if (a[i] == 1)
                    cnt1[i + 1]++;
            }
            for (int i = n; i > 0; i--) {
                cnt2[i - 1] = cnt2[i];
                if (a[i - 1] == 2)
                    cnt2[i - 1]++;
            }
            int best = 0;
            for (int i = 0; i < n; i++) {
                int dp1 = 0;
                int dp2 = 0;
                for (int j = i; j < n; j++) {
                    if (a[j] == 2)
                        dp2++;
                    else if (a[j] == 1) {
                        dp1 = Math.max(dp1 + 1, dp2 + 1);
                    }
                    best = Math.max(best, cnt1[i] + cnt2[j + 1] + Math.max(dp1, dp2));
                }
            }
            out.println(best);
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = ni();
            return a;
        }

    }
}

