#include <bits/stdc++.h>
using namespace std;
const int MAX = 3e3;
long long N, res, arr[MAX], inc[MAX][2], decr[MAX][2], dp[2][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int A = 1; A <= N; A++) cin >> arr[A];
  for (int A = 1; A <= N; A++) {
    inc[A][0] = inc[A - 1][0];
    inc[A][1] = inc[A - 1][1];
    if (arr[A] == 1)
      inc[A][0]++;
    else
      inc[A][1] = max(inc[A][1], inc[A - 1][0] + 1),
      inc[A][1] = max(inc[A][1], inc[A - 1][1] + 1);
    res = max(res, inc[A][0]);
    res = max(res, inc[A][1]);
  }
  for (int A = N; A >= 1; A--) {
    decr[A][0] = decr[A + 1][0];
    decr[A][1] = decr[A + 1][1];
    if (arr[A] == 1)
      decr[A][0] = max(decr[A + 1][0] + 1, decr[A + 1][1] + 1);
    else
      decr[A][1]++;
    res = max(res, decr[A][0]);
    res = max(res, decr[A][1]);
  }
  for (int A = N; A >= 1; A--) {
    for (int A = 0; A < 2; A++)
      for (int B = 0; B < 2; B++) dp[A][B] = 0;
    for (int B = A; B >= 1; B--) {
      if (arr[B] == 1) {
        dp[0][0]++;
        dp[1][0] = max(dp[1][0] + 1, dp[1][1] + 1);
      } else {
        dp[1][1]++;
        dp[0][1] = max(dp[0][1] + 1, dp[0][0] + 1);
      }
      res = max(res,
                dp[0][0] + inc[B - 1][0] + max(decr[A + 1][0], decr[A + 1][1]));
      res = max(res, dp[0][1] + inc[B - 1][0] + decr[A + 1][1]);
      res = max(res, dp[1][0] + inc[B - 1][0] + decr[A + 1][1]);
      res = max(res,
                dp[1][1] + max(inc[B - 1][0], inc[B - 1][1]) + decr[A + 1][1]);
    }
  }
  cout << res;
  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;
const int MAX = 3e3;
long long N, res, arr[MAX], inc[MAX][2], decr[MAX][2], dp[2][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int A = 1; A <= N; A++) cin >> arr[A];
  for (int A = 1; A <= N; A++) {
    inc[A][0] = inc[A - 1][0];
    inc[A][1] = inc[A - 1][1];
    if (arr[A] == 1)
      inc[A][0]++;
    else
      inc[A][1] = max(inc[A][1], inc[A - 1][0] + 1),
      inc[A][1] = max(inc[A][1], inc[A - 1][1] + 1);
    res = max(res, inc[A][0]);
    res = max(res, inc[A][1]);
  }
  for (int A = N; A >= 1; A--) {
    decr[A][0] = decr[A + 1][0];
    decr[A][1] = decr[A + 1][1];
    if (arr[A] == 1)
      decr[A][0] = max(decr[A + 1][0] + 1, decr[A + 1][1] + 1);
    else
      decr[A][1]++;
    res = max(res, decr[A][0]);
    res = max(res, decr[A][1]);
  }
  for (int A = N; A >= 1; A--) {
    for (int A = 0; A < 2; A++)
      for (int B = 0; B < 2; B++) dp[A][B] = 0;
    for (int B = A; B >= 1; B--) {
      if (arr[B] == 1) {
        dp[0][0]++;
        dp[1][0] = max(dp[1][0] + 1, dp[1][1] + 1);
      } else {
        dp[1][1]++;
        dp[0][1] = max(dp[0][1] + 1, dp[0][0] + 1);
      }
      res = max(res,
                dp[0][0] + inc[B - 1][0] + max(decr[A + 1][0], decr[A + 1][1]));
      res = max(res, dp[0][1] + inc[B - 1][0] + decr[A + 1][1]);
      res = max(res, dp[1][0] + inc[B - 1][0] + decr[A + 1][1]);
      res = max(res,
                dp[1][1] + max(inc[B - 1][0], inc[B - 1][1]) + decr[A + 1][1]);
    }
  }
  cout << res;
  return 0;
}

```