#include <bits/stdc++.h>
using namespace std;
int n, m, gg[2010][2][2], f[2010][2][2010][2], dp[2010][2];
bool a[2010];
int main() {
  int ans = 0;
  scanf("%d", &n);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[i] = x - 1;
  }
  for (int i = 1; i <= n; i++) {
    gg[i][0][0] = gg[i - 1][0][0] + (a[i] == 0);
    gg[i][0][1] = gg[i - 1][0][1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    gg[i][1][0] = gg[i + 1][1][0] + (a[i] == 0);
    gg[i][1][1] = gg[i + 1][1][1] + (a[i] == 1);
  }
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) f[i][a[i]][i][a[i]] = 1;
  int j;
  for (int l = 2; l <= n; l++) {
    for (int i = 1; (j = i + l - 1) <= n; i++) {
      f[i][0][j][0] = max(f[i + 1][0][j][0], f[i][0][j - 1][0]);
      f[i][1][j][0] = max(f[i + 1][1][j][0], f[i][1][j - 1][0]);
      f[i][1][j][1] = max(f[i + 1][1][j][1], f[i][1][j - 1][1]);
      if (a[i] == 0) {
        f[i][0][j][0] = max(f[i][0][j][0], f[i + 1][0][j][0] + 1);
      } else {
        f[i][1][j][0] =
            max(f[i][1][j][0], max(f[i + 1][0][j][0], f[i + 1][1][j][0]) + 1);
        f[i][1][j][1] = max(f[i][1][j][1], f[i + 1][1][j][1] + 1);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
    for (int j = i - 1; j; j--)
      if (a[i] >= a[j]) dp[i][0] = max(dp[i][0], dp[j][0] + 1);
  }
  for (int i = n; i; i--) {
    dp[i][1] = 1;
    for (int j = i + 1; j <= n; j++)
      if (a[i] <= a[j]) dp[i][1] = max(dp[i][1], dp[j][1] + 1);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans = max(ans, f[i][0][j][0] + gg[i - 1][0][0] + dp[j + 1][1]);
      ans = max(ans, f[i][1][j][0] + gg[i - 1][0][0] + gg[j + 1][1][1]);
      ans = max(ans, f[i][1][j][1] + dp[i - 1][0] + gg[j + 1][1][1]);
    }
  cout << ans << endl;
  return 0;
}

### 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 n, m, gg[2010][2][2], f[2010][2][2010][2], dp[2010][2];
bool a[2010];
int main() {
  int ans = 0;
  scanf("%d", &n);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[i] = x - 1;
  }
  for (int i = 1; i <= n; i++) {
    gg[i][0][0] = gg[i - 1][0][0] + (a[i] == 0);
    gg[i][0][1] = gg[i - 1][0][1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    gg[i][1][0] = gg[i + 1][1][0] + (a[i] == 0);
    gg[i][1][1] = gg[i + 1][1][1] + (a[i] == 1);
  }
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) f[i][a[i]][i][a[i]] = 1;
  int j;
  for (int l = 2; l <= n; l++) {
    for (int i = 1; (j = i + l - 1) <= n; i++) {
      f[i][0][j][0] = max(f[i + 1][0][j][0], f[i][0][j - 1][0]);
      f[i][1][j][0] = max(f[i + 1][1][j][0], f[i][1][j - 1][0]);
      f[i][1][j][1] = max(f[i + 1][1][j][1], f[i][1][j - 1][1]);
      if (a[i] == 0) {
        f[i][0][j][0] = max(f[i][0][j][0], f[i + 1][0][j][0] + 1);
      } else {
        f[i][1][j][0] =
            max(f[i][1][j][0], max(f[i + 1][0][j][0], f[i + 1][1][j][0]) + 1);
        f[i][1][j][1] = max(f[i][1][j][1], f[i + 1][1][j][1] + 1);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
    for (int j = i - 1; j; j--)
      if (a[i] >= a[j]) dp[i][0] = max(dp[i][0], dp[j][0] + 1);
  }
  for (int i = n; i; i--) {
    dp[i][1] = 1;
    for (int j = i + 1; j <= n; j++)
      if (a[i] <= a[j]) dp[i][1] = max(dp[i][1], dp[j][1] + 1);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans = max(ans, f[i][0][j][0] + gg[i - 1][0][0] + dp[j + 1][1]);
      ans = max(ans, f[i][1][j][0] + gg[i - 1][0][0] + gg[j + 1][1][1]);
      ans = max(ans, f[i][1][j][1] + dp[i - 1][0] + gg[j + 1][1][1]);
    }
  cout << ans << endl;
  return 0;
}

```