#include <bits/stdc++.h>
using namespace std;
int n, a[2005], st1[2005], st2[2005], dr1[2005], dr2[2005], ma;
int dp11[2005][2005], dp12[2005][2005], dp22[2005][2005], dpd2[2005][2005];
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      st1[i] = st1[i - 1] + 1;
    else
      st1[i] = st1[i - 1];
    if (a[i] == 2)
      st2[i] = max(st1[i - 1], st2[i - 1]) + 1;
    else
      st2[i] = st2[i - 1];
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 1)
      dr1[i] = max(dr1[i + 1], dr2[i + 1]) + 1;
    else
      dr1[i] = dr1[i + 1];
    if (a[i] == 2)
      dr2[i] = dr2[i + 1] + 1;
    else
      dr2[i] = dr2[i + 1];
  }
  for (int i = 1; i <= n; i++) ma = max(ma, max(st1[i], st2[i]));
  for (int d = 0; d <= n - 1; d++)
    for (int i = 1; i <= n; i++) {
      int j = i + d;
      if (j > n) break;
      if (d == 0) {
        if (a[i] == 1)
          dp11[i][i] = 1;
        else
          dp22[i][i] = 1;
      } else {
        if (a[j] == 1) {
          dp22[i][j] = dp22[i][j - 1];
          if (dp11[i][j - 1]) dp11[i][j] = 1 + dp11[i][j - 1];
          if (dp12[i][j - 1]) dp12[i][j] = 1 + dp12[i][j - 1];
          if (dp22[i][j - 1]) dp12[i][j] = max(dp12[i][j], 1 + dp22[i][j - 1]);
        }
        if (a[j] == 2) {
          dp11[i][j] = dp11[i][j - 1];
          dp12[i][j] = dp12[i][j - 1];
          if (dp22[i][j - 1]) dp22[i][j] = 1 + dp22[i][j - 1];
        }
      }
      ma = max(ma, st1[i - 1] + dp11[i][j] + max(dr1[j + 1], dr2[j + 1]));
      ma = max(ma, st1[i - 1] + dp12[i][j] + dr2[j + 1]);
      ma = max(ma, max(st1[i - 1], st2[i - 1]) + dp22[i][j] + dr2[j + 1]);
    }
  cout << ma;
}

### Prompt

Please formulate 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], st1[2005], st2[2005], dr1[2005], dr2[2005], ma;
int dp11[2005][2005], dp12[2005][2005], dp22[2005][2005], dpd2[2005][2005];
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      st1[i] = st1[i - 1] + 1;
    else
      st1[i] = st1[i - 1];
    if (a[i] == 2)
      st2[i] = max(st1[i - 1], st2[i - 1]) + 1;
    else
      st2[i] = st2[i - 1];
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 1)
      dr1[i] = max(dr1[i + 1], dr2[i + 1]) + 1;
    else
      dr1[i] = dr1[i + 1];
    if (a[i] == 2)
      dr2[i] = dr2[i + 1] + 1;
    else
      dr2[i] = dr2[i + 1];
  }
  for (int i = 1; i <= n; i++) ma = max(ma, max(st1[i], st2[i]));
  for (int d = 0; d <= n - 1; d++)
    for (int i = 1; i <= n; i++) {
      int j = i + d;
      if (j > n) break;
      if (d == 0) {
        if (a[i] == 1)
          dp11[i][i] = 1;
        else
          dp22[i][i] = 1;
      } else {
        if (a[j] == 1) {
          dp22[i][j] = dp22[i][j - 1];
          if (dp11[i][j - 1]) dp11[i][j] = 1 + dp11[i][j - 1];
          if (dp12[i][j - 1]) dp12[i][j] = 1 + dp12[i][j - 1];
          if (dp22[i][j - 1]) dp12[i][j] = max(dp12[i][j], 1 + dp22[i][j - 1]);
        }
        if (a[j] == 2) {
          dp11[i][j] = dp11[i][j - 1];
          dp12[i][j] = dp12[i][j - 1];
          if (dp22[i][j - 1]) dp22[i][j] = 1 + dp22[i][j - 1];
        }
      }
      ma = max(ma, st1[i - 1] + dp11[i][j] + max(dr1[j + 1], dr2[j + 1]));
      ma = max(ma, st1[i - 1] + dp12[i][j] + dr2[j + 1]);
      ma = max(ma, max(st1[i - 1], st2[i - 1]) + dp22[i][j] + dr2[j + 1]);
    }
  cout << ma;
}

```