#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
int arr[N];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n == 1) {
    cout << 1;
    return 0;
  }
  int cnt2 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 2) {
      cnt2++;
    }
  }
  int fin_result = 0;
  int tcnt1 = 0, tcnt2 = 0;
  for (int i = 0; i < n; i++) {
    tcnt1 = tcnt2 = 0;
    for (int j = 0; j < i; j++) {
      if (arr[j] == 1) tcnt1++;
    }
    for (int j = i + 1; j < n; j++) {
      if (arr[j] == 2) tcnt2++;
    }
    fin_result = max(fin_result, tcnt1 + tcnt2 + 1);
  }
  int cnt1 = 0;
  int dp_1 = 0, dp_2 = 0;
  for (int i = 0; i < n; i++) {
    dp_1 = dp_2 = 0;
    int temp_result = 0;
    if (arr[i] == 2) {
      dp_2 = 1;
      cnt2--;
    } else if (arr[i] == 1) {
      dp_1 = 1;
    }
    int tmp = cnt2;
    for (int j = i + 1; j < n; j++) {
      if (arr[j] == 1) {
        dp_1 = max(dp_1, dp_2) + 1;
      } else if (arr[j] == 2) {
        dp_2++;
        tmp--;
      }
      temp_result = max(temp_result, cnt1 + max(dp_1, dp_2) + tmp);
    }
    if (arr[i] == 1) {
      cnt1++;
    }
    fin_result = max(fin_result, temp_result);
  }
  cout << fin_result << "\n";
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 2010;
int arr[N];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n == 1) {
    cout << 1;
    return 0;
  }
  int cnt2 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 2) {
      cnt2++;
    }
  }
  int fin_result = 0;
  int tcnt1 = 0, tcnt2 = 0;
  for (int i = 0; i < n; i++) {
    tcnt1 = tcnt2 = 0;
    for (int j = 0; j < i; j++) {
      if (arr[j] == 1) tcnt1++;
    }
    for (int j = i + 1; j < n; j++) {
      if (arr[j] == 2) tcnt2++;
    }
    fin_result = max(fin_result, tcnt1 + tcnt2 + 1);
  }
  int cnt1 = 0;
  int dp_1 = 0, dp_2 = 0;
  for (int i = 0; i < n; i++) {
    dp_1 = dp_2 = 0;
    int temp_result = 0;
    if (arr[i] == 2) {
      dp_2 = 1;
      cnt2--;
    } else if (arr[i] == 1) {
      dp_1 = 1;
    }
    int tmp = cnt2;
    for (int j = i + 1; j < n; j++) {
      if (arr[j] == 1) {
        dp_1 = max(dp_1, dp_2) + 1;
      } else if (arr[j] == 2) {
        dp_2++;
        tmp--;
      }
      temp_result = max(temp_result, cnt1 + max(dp_1, dp_2) + tmp);
    }
    if (arr[i] == 1) {
      cnt1++;
    }
    fin_result = max(fin_result, temp_result);
  }
  cout << fin_result << "\n";
}

```