#include <bits/stdc++.h>
using namespace std;
int m, num[2005], dp[2][2005], find[2005][2005];
int main() {
  cin >> m;
  for (int i = 1; i <= m; i++) scanf("%d", &num[i]);
  dp[0][0] = dp[1][m + 1] = 0;
  for (int i = 1; i <= m; i++) dp[0][i] = dp[0][i - 1] + (num[i] == 1 ? 1 : 0);
  for (int i = m; i >= 1; i--) dp[1][i] = dp[1][i + 1] + (num[i] == 2 ? 1 : 0);
  int len = 0;
  for (int i = 1; i <= m; i++) len = max(len, dp[0][i] + dp[1][i]);
  for (int i = 1, j1 = 0, j2 = 0; i <= m; i++, j1 = 0, j2 = 0)
    for (int j = i; j <= m; j++) {
      if (num[j] == 2)
        j2++, j1 ? j1-- : j1;
      else
        j1++;
      len = max(len, dp[0][i - 1] + j1 + j2 + dp[1][j + 1]);
    }
  cout << len << endl;
  return 0;
}

### 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;
int m, num[2005], dp[2][2005], find[2005][2005];
int main() {
  cin >> m;
  for (int i = 1; i <= m; i++) scanf("%d", &num[i]);
  dp[0][0] = dp[1][m + 1] = 0;
  for (int i = 1; i <= m; i++) dp[0][i] = dp[0][i - 1] + (num[i] == 1 ? 1 : 0);
  for (int i = m; i >= 1; i--) dp[1][i] = dp[1][i + 1] + (num[i] == 2 ? 1 : 0);
  int len = 0;
  for (int i = 1; i <= m; i++) len = max(len, dp[0][i] + dp[1][i]);
  for (int i = 1, j1 = 0, j2 = 0; i <= m; i++, j1 = 0, j2 = 0)
    for (int j = i; j <= m; j++) {
      if (num[j] == 2)
        j2++, j1 ? j1-- : j1;
      else
        j1++;
      len = max(len, dp[0][i - 1] + j1 + j2 + dp[1][j + 1]);
    }
  cout << len << endl;
  return 0;
}

```