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.
//package codeforces.r934;

import java.util.Scanner;

public class C {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    int a[] = new int[n];
    for (int i = 0; i < n; i++) {
      a[i] = scanner.nextInt();
    }

    int cnt1[] = new int[n];
    int cnt2[] = new int[n];
    int mm[] = new int[n];

    cnt1[0] = a[0] == 1 ? 1 : 0;
    for (int i = 1; i < n; i++) {
      cnt1[i] = a[i] == 1? cnt1[i-1] + 1 : cnt1[i-1];
    }

    cnt2[n-1] = a[n-1] == 2 ? 1 : 0;
    for (int i = n-2; i >= 0; i--) {
      cnt2[i] = a[i] == 2? cnt2[i+1] + 1 : cnt2[i+1];
    }

    for (int i = 0; i < n; i++) {
      int left = i == 0 ? 0 : cnt1[i-1];
      int cnt = 0;
      for (int j = i-1; j >= 0; j--) {
        if (a[j] == 2) cnt ++;
        int ll = j == 0 ? 0 : cnt1[j-1];
        left = Math.max(left, ll + cnt);
      }

      int right = i == n-1 ? 0 : cnt2[i+1];
      cnt = 0;
      for (int j = i+1; j < n; j++) {
        if (a[j] == 1) cnt ++;
        int rr = j == n-1 ? 0 : cnt2[j+1];
        right = Math.max(right, rr + cnt);
      }
      mm[i] = left + right + 1;
    }

    int max = mm[0];
    for (int i = 0; i < n; i++) {
      max = Math.max(max, mm[i]);
    }

    System.out.printf("%d\n", max);

  }
}