#include <bits/stdc++.h>
using namespace std;
int o[2005], t[2005];
int to[2005][2005][2];
int a[2005];
int n;
void solve() {
  for (int i = 1; i <= n; i++) {
    o[i] = o[i - 1] + ((a[i] == 1) ? 1 : 0);
  }
  for (int j = n; j >= 1; j--) {
    t[j] = t[j + 1] + ((a[j] == 2) ? 1 : 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (a[j] == 2) {
        to[i][j][1] = to[i][j - 1][1] + 1;
        to[i][j][0] = to[i][j - 1][0];
      } else {
        to[i][j][0] = max(to[i][j - 1][0] + 1, to[i][j - 1][1] + 1);
        to[i][j][1] = to[i][j - 1][1];
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(o[i - 1] + max(to[i][j][0], to[i][j][1]) + t[j + 1], ans);
    }
  }
  cout << ans << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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 o[2005], t[2005];
int to[2005][2005][2];
int a[2005];
int n;
void solve() {
  for (int i = 1; i <= n; i++) {
    o[i] = o[i - 1] + ((a[i] == 1) ? 1 : 0);
  }
  for (int j = n; j >= 1; j--) {
    t[j] = t[j + 1] + ((a[j] == 2) ? 1 : 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (a[j] == 2) {
        to[i][j][1] = to[i][j - 1][1] + 1;
        to[i][j][0] = to[i][j - 1][0];
      } else {
        to[i][j][0] = max(to[i][j - 1][0] + 1, to[i][j - 1][1] + 1);
        to[i][j][1] = to[i][j - 1][1];
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(o[i - 1] + max(to[i][j][0], to[i][j][1]) + t[j + 1], ans);
    }
  }
  cout << ans << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  solve();
  return 0;
}

```