#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int dp[n][4];
  memset(dp, 0, sizeof(dp));
  if (arr[0] == 1) {
    dp[0][0] = 1;
    dp[0][2] = 1;
  } else {
    dp[0][1] = 1;
    dp[0][3] = 1;
  }
  for (int i = 1; i < n; i++) {
    if (arr[i] == 1) {
      dp[i][0] = dp[i - 1][0] + 1;
      dp[i][1] = dp[i - 1][1];
      dp[i][2] = max(dp[i - 1][1], dp[i - 1][2]) + 1;
      dp[i][3] = dp[i - 1][3];
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + 1;
      dp[i][2] = dp[i - 1][2];
      dp[i][3] = max(dp[i - 1][2], dp[i - 1][3]) + 1;
    }
  }
  cout << max(dp[n - 1][2], dp[n - 1][3]);
}

### Prompt

Create a solution in CPP for 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 main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int dp[n][4];
  memset(dp, 0, sizeof(dp));
  if (arr[0] == 1) {
    dp[0][0] = 1;
    dp[0][2] = 1;
  } else {
    dp[0][1] = 1;
    dp[0][3] = 1;
  }
  for (int i = 1; i < n; i++) {
    if (arr[i] == 1) {
      dp[i][0] = dp[i - 1][0] + 1;
      dp[i][1] = dp[i - 1][1];
      dp[i][2] = max(dp[i - 1][1], dp[i - 1][2]) + 1;
      dp[i][3] = dp[i - 1][3];
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + 1;
      dp[i][2] = dp[i - 1][2];
      dp[i][3] = max(dp[i - 1][2], dp[i - 1][3]) + 1;
    }
  }
  cout << max(dp[n - 1][2], dp[n - 1][3]);
}

```