#include <bits/stdc++.h>
using namespace std;
int arr[2002];
int mx[4];
int best[2002][2002];
int main() {
  int n;
  scanf("%d", &n);
  int allt = 0, allo = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", arr + i);
    if (arr[i] == 2)
      ++allt;
    else
      allo++;
  }
  for (int j = n - 1; j >= 0; --j) {
    memset(mx, 0, sizeof mx);
    for (int i = j; i >= 0; --i) {
      if (arr[i] == 1) {
        mx[1]++;
      } else {
        mx[2] = max(mx[2] + 1, 1 + mx[1]);
      }
      best[i][j] = mx[2];
    }
  }
  int ans = 0;
  int bo = 0, bt = 0;
  for (int i = 0; i < n; ++i) {
    int tw = 0;
    for (int j = i; j < n; ++j) {
      if (arr[j] == 2) tw++;
      ans = max(ans, bo + (allt - bt - tw) + best[i][j]);
    }
    if (arr[i] == 2)
      bt++;
    else
      bo++;
  }
  ans = max(ans, max(allo, allt));
  printf("%d", ans);
}

### 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 arr[2002];
int mx[4];
int best[2002][2002];
int main() {
  int n;
  scanf("%d", &n);
  int allt = 0, allo = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", arr + i);
    if (arr[i] == 2)
      ++allt;
    else
      allo++;
  }
  for (int j = n - 1; j >= 0; --j) {
    memset(mx, 0, sizeof mx);
    for (int i = j; i >= 0; --i) {
      if (arr[i] == 1) {
        mx[1]++;
      } else {
        mx[2] = max(mx[2] + 1, 1 + mx[1]);
      }
      best[i][j] = mx[2];
    }
  }
  int ans = 0;
  int bo = 0, bt = 0;
  for (int i = 0; i < n; ++i) {
    int tw = 0;
    for (int j = i; j < n; ++j) {
      if (arr[j] == 2) tw++;
      ans = max(ans, bo + (allt - bt - tw) + best[i][j]);
    }
    if (arr[i] == 2)
      bt++;
    else
      bo++;
  }
  ans = max(ans, max(allo, allt));
  printf("%d", ans);
}

```