#include <bits/stdc++.h>
using namespace std;
const long long lim = 2e5;
int main() {
  long long m, n, A[lim] = {}, B[lim], C[lim], M;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> A[i];
  B[0] = 0;
  for (long long i = 1; i < n; i++) {
    if (A[i - 1] == 1)
      B[i] = B[i - 1] + 1;
    else
      B[i] = B[i - 1];
  }
  C[n - 1] = 0;
  for (long long i = n - 2; i >= 0; i--) {
    if (A[i + 1] == 2)
      C[i] = C[i + 1] + 1;
    else
      C[i] = C[i + 1];
  }
  long long Ans = 0;
  for (long long i = 0; i < n; i++) {
    long long endwith1 = 0, endwith2 = 0;
    for (long long j = i; j >= 0; j--) {
      if (A[j] == 2) {
        endwith2 = max(endwith2 + 1, endwith1 + 1);
      } else {
        endwith1++;
      }
      Ans = max(Ans, B[j] + endwith1 + C[i]);
      Ans = max(Ans, B[j] + endwith2 + C[i]);
    }
  }
  for (long long i = 0; i < n; i++) {
    Ans = max(Ans, B[i] + C[i] + 1);
  }
  cout << Ans << endl;
  return 0;
}

### 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;
const long long lim = 2e5;
int main() {
  long long m, n, A[lim] = {}, B[lim], C[lim], M;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> A[i];
  B[0] = 0;
  for (long long i = 1; i < n; i++) {
    if (A[i - 1] == 1)
      B[i] = B[i - 1] + 1;
    else
      B[i] = B[i - 1];
  }
  C[n - 1] = 0;
  for (long long i = n - 2; i >= 0; i--) {
    if (A[i + 1] == 2)
      C[i] = C[i + 1] + 1;
    else
      C[i] = C[i + 1];
  }
  long long Ans = 0;
  for (long long i = 0; i < n; i++) {
    long long endwith1 = 0, endwith2 = 0;
    for (long long j = i; j >= 0; j--) {
      if (A[j] == 2) {
        endwith2 = max(endwith2 + 1, endwith1 + 1);
      } else {
        endwith1++;
      }
      Ans = max(Ans, B[j] + endwith1 + C[i]);
      Ans = max(Ans, B[j] + endwith2 + C[i]);
    }
  }
  for (long long i = 0; i < n; i++) {
    Ans = max(Ans, B[i] + C[i] + 1);
  }
  cout << Ans << endl;
  return 0;
}

```