#include <bits/stdc++.h>
using namespace std;
int ar[2005], lis[2005][2005], on[2005], tw[2005];
multiset<int> s;
multiset<int>::iterator it;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    on[i] = on[i - 1] + (ar[i] == 1);
    tw[i] = tw[i - 1] + (ar[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    s.clear();
    s.insert(ar[i]);
    lis[i][i] = (int)s.size();
    for (int j = i - 1; j >= 1; j--) {
      s.insert(ar[j]);
      it = s.upper_bound(ar[j]);
      if (it != s.end()) s.erase(it);
      lis[j][i] = (int)s.size();
    }
  }
  int ans = (int)s.size();
  ans = max(ans, max(on[n], tw[n]));
  for (int i = 1; i <= n - 1; i++) {
    for (int j = i; j <= n; j++) {
      int l = i, r = j;
      int x = lis[l][r];
      int o = on[l - 1];
      int t = tw[n] - tw[r];
      x += (o + t);
      ans = max(ans, x);
    }
  }
  cout << ans << 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 ar[2005], lis[2005][2005], on[2005], tw[2005];
multiset<int> s;
multiset<int>::iterator it;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    on[i] = on[i - 1] + (ar[i] == 1);
    tw[i] = tw[i - 1] + (ar[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    s.clear();
    s.insert(ar[i]);
    lis[i][i] = (int)s.size();
    for (int j = i - 1; j >= 1; j--) {
      s.insert(ar[j]);
      it = s.upper_bound(ar[j]);
      if (it != s.end()) s.erase(it);
      lis[j][i] = (int)s.size();
    }
  }
  int ans = (int)s.size();
  ans = max(ans, max(on[n], tw[n]));
  for (int i = 1; i <= n - 1; i++) {
    for (int j = i; j <= n; j++) {
      int l = i, r = j;
      int x = lis[l][r];
      int o = on[l - 1];
      int t = tw[n] - tw[r];
      x += (o + t);
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
  return 0;
}

```