#include <bits/stdc++.h>
using namespace std;
int dp[2002][2002][2][2];
int main() {
  int n;
  scanf("%d", &n);
  int arr[n + 2];
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  for (int j = 1; j <= n; ++j) {
    for (int k = 0; k < 2; ++k) {
      dp[j][j][k][k] = arr[j] == k + 1 ? 1 : 0;
    }
  }
  for (int j = 1; j <= n; ++j) {
    for (int i = 1; i <= n; ++i) {
      for (int startNumber = 0; startNumber < 2 && i != j; ++startNumber) {
        for (int endNumber = 0; endNumber < 2; ++endNumber) {
          if (!startNumber && !endNumber)
            dp[j][i][startNumber][endNumber] =
                dp[j][i - 1][startNumber][endNumber] + (arr[i] == 1 ? 1 : 0);
          if (!startNumber && endNumber) {
            dp[j][i][startNumber][endNumber] =
                max(dp[j][i - 1][startNumber][endNumber],
                    dp[j][i - 1][startNumber][0]) +
                (arr[i] == 2 ? 1 : 0);
          }
          if (startNumber && !endNumber) {
            dp[j][i][startNumber][endNumber] =
                max(dp[j][i - 1][1][endNumber] + (arr[i] == 1 ? 1 : 0),
                    (arr[i] == 1 ? 1 + dp[j][i - 1][1][1] : 0));
          }
          if (startNumber && endNumber) {
            dp[j][i][startNumber][endNumber] =
                dp[j][i - 1][startNumber][endNumber] + (arr[i] == 2 ? 1 : 0);
          }
        }
      }
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int temp = dp[1][l - 1][0][0];
      int temp2 =
          max(dp[l][r][0][0] + max(dp[r + 1][n][1][1], dp[r + 1][n][0][0]),
              r != l ? dp[l][r][1][0] + dp[r + 1][n][1][1] : 0);
      ans = max(ans, temp + temp2);
      temp = dp[1][l - 1][0][1] + dp[l][r][1][1] + dp[r + 1][n][1][1];
      ans = max(ans, temp);
      temp = dp[1][l - 1][1][1] + dp[l][r][1][1] + dp[r + 1][n][1][1];
      ans = max(ans, temp);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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 dp[2002][2002][2][2];
int main() {
  int n;
  scanf("%d", &n);
  int arr[n + 2];
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  for (int j = 1; j <= n; ++j) {
    for (int k = 0; k < 2; ++k) {
      dp[j][j][k][k] = arr[j] == k + 1 ? 1 : 0;
    }
  }
  for (int j = 1; j <= n; ++j) {
    for (int i = 1; i <= n; ++i) {
      for (int startNumber = 0; startNumber < 2 && i != j; ++startNumber) {
        for (int endNumber = 0; endNumber < 2; ++endNumber) {
          if (!startNumber && !endNumber)
            dp[j][i][startNumber][endNumber] =
                dp[j][i - 1][startNumber][endNumber] + (arr[i] == 1 ? 1 : 0);
          if (!startNumber && endNumber) {
            dp[j][i][startNumber][endNumber] =
                max(dp[j][i - 1][startNumber][endNumber],
                    dp[j][i - 1][startNumber][0]) +
                (arr[i] == 2 ? 1 : 0);
          }
          if (startNumber && !endNumber) {
            dp[j][i][startNumber][endNumber] =
                max(dp[j][i - 1][1][endNumber] + (arr[i] == 1 ? 1 : 0),
                    (arr[i] == 1 ? 1 + dp[j][i - 1][1][1] : 0));
          }
          if (startNumber && endNumber) {
            dp[j][i][startNumber][endNumber] =
                dp[j][i - 1][startNumber][endNumber] + (arr[i] == 2 ? 1 : 0);
          }
        }
      }
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int temp = dp[1][l - 1][0][0];
      int temp2 =
          max(dp[l][r][0][0] + max(dp[r + 1][n][1][1], dp[r + 1][n][0][0]),
              r != l ? dp[l][r][1][0] + dp[r + 1][n][1][1] : 0);
      ans = max(ans, temp + temp2);
      temp = dp[1][l - 1][0][1] + dp[l][r][1][1] + dp[r + 1][n][1][1];
      ans = max(ans, temp);
      temp = dp[1][l - 1][1][1] + dp[l][r][1][1] + dp[r + 1][n][1][1];
      ans = max(ans, temp);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```