#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 2, K = 19;
const int mod = 1e9 + 7;
int arr[N], rv[2020][2020][3], nx[2020][2020][3];
int max(int a, int b, int c) { return max(max(a, b), c); }
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    arr[i]--;
  }
  int mx = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      rv[i + 1][j + 1][0] = !arr[j] + rv[i + 1][j][0];
      rv[i + 1][j + 1][1] = arr[j] + rv[i + 1][j][1];
      rv[i + 1][j + 1][2] =
          max(!arr[j] + rv[i + 1][j][0], arr[j] + rv[i + 1][j][2]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j >= 0; j--) {
      nx[j + 1][i + 1][0] = !arr[j] + nx[j + 2][i + 1][0];
      nx[j + 1][i + 1][1] = arr[j] + nx[j + 2][i + 1][1];
      nx[j + 1][i + 1][2] =
          max(!arr[j] + nx[j + 2][i + 1][0], arr[j] + nx[j + 2][i + 1][2]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int a = rv[1][i][2] + nx[i + 1][j + 1][1] + rv[j + 2][n][1];
      int b = rv[1][i][0] + nx[i + 1][j + 1][2] + rv[j + 2][n][1];
      int c = rv[1][i][0] + nx[i + 1][j + 1][0] + rv[j + 2][n][2];
      mx = max(mx, max(a, b, c));
    }
  }
  cout << mx;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int N = 2e5 + 2, K = 19;
const int mod = 1e9 + 7;
int arr[N], rv[2020][2020][3], nx[2020][2020][3];
int max(int a, int b, int c) { return max(max(a, b), c); }
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    arr[i]--;
  }
  int mx = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      rv[i + 1][j + 1][0] = !arr[j] + rv[i + 1][j][0];
      rv[i + 1][j + 1][1] = arr[j] + rv[i + 1][j][1];
      rv[i + 1][j + 1][2] =
          max(!arr[j] + rv[i + 1][j][0], arr[j] + rv[i + 1][j][2]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j >= 0; j--) {
      nx[j + 1][i + 1][0] = !arr[j] + nx[j + 2][i + 1][0];
      nx[j + 1][i + 1][1] = arr[j] + nx[j + 2][i + 1][1];
      nx[j + 1][i + 1][2] =
          max(!arr[j] + nx[j + 2][i + 1][0], arr[j] + nx[j + 2][i + 1][2]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int a = rv[1][i][2] + nx[i + 1][j + 1][1] + rv[j + 2][n][1];
      int b = rv[1][i][0] + nx[i + 1][j + 1][2] + rv[j + 2][n][1];
      int c = rv[1][i][0] + nx[i + 1][j + 1][0] + rv[j + 2][n][2];
      mx = max(mx, max(a, b, c));
    }
  }
  cout << mx;
  return 0;
}

```