#include <bits/stdc++.h>
using namespace std;
int a[20005];
int one_count[20005], two_count[20005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    cin >> a[i];
    one_count[i] = 0;
    two_count[i] = 0;
    if (i == 0) {
      if (a[i] == 1)
        one_count[0] = 1;
      else
        two_count[0] = 1;
    }
  }
  for (__typeof((n)) i = (1); i < (n); i += 1) {
    one_count[i] = one_count[i - 1];
    two_count[i] = two_count[i - 1];
    if (a[i] == 1)
      one_count[i]++;
    else
      two_count[i]++;
  }
  int total_two = two_count[n - 1];
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    two_count[i] = total_two - two_count[i];
  }
  int normal = total_two;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    normal = ((normal) > (one_count[i] + two_count[i])
                  ? (normal)
                  : (one_count[i] + two_count[i]));
  }
  int ans = normal;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    int two_ending = 0, one_ending = 0;
    if (a[i] == 2) {
      two_ending = 1;
    } else {
      one_ending = 1;
    }
    for (__typeof((n)) j = (i + 1); j < (n); j += 1) {
      int temp_ans = 0;
      if (i != 0)
        temp_ans = one_count[i - 1] + two_count[j];
      else
        temp_ans = two_count[j];
      if (a[j] == 1) {
        one_ending = ((one_ending + 1) > (two_ending + 1) ? (one_ending + 1)
                                                          : (two_ending + 1));
      } else {
        two_ending++;
      }
      int longest = ((one_ending) > (two_ending) ? (one_ending) : (two_ending));
      temp_ans += longest;
      ans = ((ans) > (temp_ans) ? (ans) : (temp_ans));
    }
  }
  cout << ans;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[20005];
int one_count[20005], two_count[20005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    cin >> a[i];
    one_count[i] = 0;
    two_count[i] = 0;
    if (i == 0) {
      if (a[i] == 1)
        one_count[0] = 1;
      else
        two_count[0] = 1;
    }
  }
  for (__typeof((n)) i = (1); i < (n); i += 1) {
    one_count[i] = one_count[i - 1];
    two_count[i] = two_count[i - 1];
    if (a[i] == 1)
      one_count[i]++;
    else
      two_count[i]++;
  }
  int total_two = two_count[n - 1];
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    two_count[i] = total_two - two_count[i];
  }
  int normal = total_two;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    normal = ((normal) > (one_count[i] + two_count[i])
                  ? (normal)
                  : (one_count[i] + two_count[i]));
  }
  int ans = normal;
  for (__typeof((n)) i = (0); i < (n); i += 1) {
    int two_ending = 0, one_ending = 0;
    if (a[i] == 2) {
      two_ending = 1;
    } else {
      one_ending = 1;
    }
    for (__typeof((n)) j = (i + 1); j < (n); j += 1) {
      int temp_ans = 0;
      if (i != 0)
        temp_ans = one_count[i - 1] + two_count[j];
      else
        temp_ans = two_count[j];
      if (a[j] == 1) {
        one_ending = ((one_ending + 1) > (two_ending + 1) ? (one_ending + 1)
                                                          : (two_ending + 1));
      } else {
        two_ending++;
      }
      int longest = ((one_ending) > (two_ending) ? (one_ending) : (two_ending));
      temp_ans += longest;
      ans = ((ans) > (temp_ans) ? (ans) : (temp_ans));
    }
  }
  cout << ans;
}

```