#include <bits/stdc++.h>
using namespace std;
int n, a[2005], f1[2005], f2[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  int cur = 0, res = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) cur++;
    f1[i] = cur;
  }
  cur = 0;
  for (int i = n; i >= 1; --i) {
    if (a[i] == 2) cur++;
    f2[i] = cur;
  }
  cur = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      cur++;
      res = max(res, cur);
    } else
      res++;
  }
  int cnt;
  for (int i = 1; i <= n; ++i) {
    cur = cnt = 0;
    for (int j = i; j >= 1; --j) {
      if (a[j] == 1) {
        cur++;
        cnt = max(cnt, cur);
      } else
        cnt++;
      if (i - j + 1 > 1) {
        res = max(res, cnt + f1[j - 1] + f2[i + 1]);
      }
    }
  }
  cout << res;
}

### 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, a[2005], f1[2005], f2[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  int cur = 0, res = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) cur++;
    f1[i] = cur;
  }
  cur = 0;
  for (int i = n; i >= 1; --i) {
    if (a[i] == 2) cur++;
    f2[i] = cur;
  }
  cur = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      cur++;
      res = max(res, cur);
    } else
      res++;
  }
  int cnt;
  for (int i = 1; i <= n; ++i) {
    cur = cnt = 0;
    for (int j = i; j >= 1; --j) {
      if (a[j] == 1) {
        cur++;
        cnt = max(cnt, cur);
      } else
        cnt++;
      if (i - j + 1 > 1) {
        res = max(res, cnt + f1[j - 1] + f2[i + 1]);
      }
    }
  }
  cout << res;
}

```