#include <bits/stdc++.h>
using namespace std;
int num[2010][2010][3];
int arr[2010];
int out[2010][2010][3];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> arr[i];
    arr[i]--;
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (i); j < (n); j++) {
      if (j) {
        num[i][j][0] = num[i][j - 1][0];
        num[i][j][1] = num[i][j - 1][1];
      }
      num[i][j][0] += (arr[j] == 0);
      num[i][j][1] += (arr[j] == 1);
    }
  }
  for (int i = (0); i < (n); i++) {
    out[i][i][0] = num[i][i][1];
    out[i][i][1] = num[i][i][0];
    for (int j = (i + 1); j < (n); j++) {
      out[i][j][0] = out[i][j - 1][0] + (arr[j] == 1);
      out[i][j][1] = out[i][j - 1][1] + (arr[j] == 0);
      out[i][j][1] = max(out[i][j][1], out[i][j - 1][0] + (arr[j] == 0));
    }
  }
  int sol = max(out[0][n - 1][0], out[0][n - 1][1]);
  for (int i = (0); i < (n); i++) {
    for (int j = (i); j < (n); j++) {
      sol = max(sol, num[0][i][0] + num[i][n - 1][1]);
    }
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (i + 2); j < (n); j++) {
      sol = max(sol, num[0][i][0] + num[j][n - 1][1] +
                         max(out[i + 1][j - 1][0], out[i + 1][j - 1][1]));
    }
  }
  for (int i = (0); i < (n - 1); i++) {
    sol = max(sol,
              num[0][i][0] + max(out[i + 1][n - 1][0], out[i + 1][n - 1][1]));
  }
  for (int i = (1); i < (n); i++) {
    sol = max(sol, num[i][n - 1][1] + max(out[0][i - 1][0], out[0][i - 1][1]));
  }
  cout << sol << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 num[2010][2010][3];
int arr[2010];
int out[2010][2010][3];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> arr[i];
    arr[i]--;
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (i); j < (n); j++) {
      if (j) {
        num[i][j][0] = num[i][j - 1][0];
        num[i][j][1] = num[i][j - 1][1];
      }
      num[i][j][0] += (arr[j] == 0);
      num[i][j][1] += (arr[j] == 1);
    }
  }
  for (int i = (0); i < (n); i++) {
    out[i][i][0] = num[i][i][1];
    out[i][i][1] = num[i][i][0];
    for (int j = (i + 1); j < (n); j++) {
      out[i][j][0] = out[i][j - 1][0] + (arr[j] == 1);
      out[i][j][1] = out[i][j - 1][1] + (arr[j] == 0);
      out[i][j][1] = max(out[i][j][1], out[i][j - 1][0] + (arr[j] == 0));
    }
  }
  int sol = max(out[0][n - 1][0], out[0][n - 1][1]);
  for (int i = (0); i < (n); i++) {
    for (int j = (i); j < (n); j++) {
      sol = max(sol, num[0][i][0] + num[i][n - 1][1]);
    }
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (i + 2); j < (n); j++) {
      sol = max(sol, num[0][i][0] + num[j][n - 1][1] +
                         max(out[i + 1][j - 1][0], out[i + 1][j - 1][1]));
    }
  }
  for (int i = (0); i < (n - 1); i++) {
    sol = max(sol,
              num[0][i][0] + max(out[i + 1][n - 1][0], out[i + 1][n - 1][1]));
  }
  for (int i = (1); i < (n); i++) {
    sol = max(sol, num[i][n - 1][1] + max(out[0][i - 1][0], out[0][i - 1][1]));
  }
  cout << sol << endl;
  return 0;
}

```