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.*;
import java.util.*;

public class Solution {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    static final int Mod = 1000000007;

    void solve() throws IOException {
        int n = nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        int ones[] = new int[n + 1];
        int twos[] = new int[n + 1];
        for (int i = 0; i < n; i++) {
            ones[i + 1] = ones[i];
            twos[i + 1] = twos[i];
            if (a[i] == 1) {
                ones[i + 1]++;
            } else twos[i + 1]++;
        }
        int ans = 0;
        for (int i = 0; i <= n; i++) {
            ans = Math.max(ans, ones[i] + twos[n] - twos[i]);
        }
        for (int r = 0; r < n; r++) {
            int cur = 0;
            for (int l = r; l >= 0; l--) {
                if (a[l] == 2) cur++;
                cur = Math.max(cur, ones[r + 1] - ones[l]);
                ans = Math.max(ans, ones[l] + cur + twos[n] - twos[r + 1]);
            }
        }
        out.println(ans);
    }

    Solution() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}