#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
int arr[2000];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  int sum2 = 0;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 2) {
      ++sum2;
    }
  }
  int ans = sum2;
  for (int i = 0, cnt1 = 0, cnt2 = 0; i < n; ++i) {
    if (arr[i] == 1) {
      ++cnt1;
    } else {
      ++cnt2;
    }
    ans = max(ans, cnt1 + sum2 - cnt2);
  }
  for (int l = 0, pre1 = 0, pre2 = 0; l < n; ++l) {
    if (arr[l] == 1) {
      ++pre1;
      continue;
    }
    for (int r = l + 1, cur = 0, cnt2 = 0; r < n; ++r) {
      if (arr[r] == 2) {
        ++cnt2;
        continue;
      }
      cur = max(cur, cnt2) + 1;
      ans = max(ans, pre1 + cur + sum2 - pre2 - cnt2);
    }
    ++pre2;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
using ll = int64_t;
int arr[2000];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  int sum2 = 0;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 2) {
      ++sum2;
    }
  }
  int ans = sum2;
  for (int i = 0, cnt1 = 0, cnt2 = 0; i < n; ++i) {
    if (arr[i] == 1) {
      ++cnt1;
    } else {
      ++cnt2;
    }
    ans = max(ans, cnt1 + sum2 - cnt2);
  }
  for (int l = 0, pre1 = 0, pre2 = 0; l < n; ++l) {
    if (arr[l] == 1) {
      ++pre1;
      continue;
    }
    for (int r = l + 1, cur = 0, cnt2 = 0; r < n; ++r) {
      if (arr[r] == 2) {
        ++cnt2;
        continue;
      }
      cur = max(cur, cnt2) + 1;
      ans = max(ans, pre1 + cur + sum2 - pre2 - cnt2);
    }
    ++pre2;
  }
  cout << ans << "\n";
  return 0;
}

```