#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2000;
int a, b, c, m, n, A[1 + MAXN], C[3][1 + MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
  }
  for (int i = 1; i <= n; ++i) {
    C[1][i] += C[1][i - 1];
    C[2][i] += C[2][i - 1];
    ++C[A[i]][i];
  }
  m = max(C[1][n], C[2][n]);
  for (int i = 1; i < n; ++i) {
    m = max(m, C[1][i] + C[2][n] - C[2][i]);
  }
  for (int i = 1; i < n; ++i) {
    a = 0;
    for (int j = i - 1; j >= 0; --j) {
      c = C[1][j] + C[2][i] - C[2][j];
      a = max(a, c);
    }
    b = 0;
    for (int j = i + 1; j <= n; ++j) {
      c = C[1][j] - C[1][i] + C[2][n] - C[2][j];
      b = max(b, c);
    }
    m = max(m, a + b);
  }
  cout << m << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int MAXN = 2000;
int a, b, c, m, n, A[1 + MAXN], C[3][1 + MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
  }
  for (int i = 1; i <= n; ++i) {
    C[1][i] += C[1][i - 1];
    C[2][i] += C[2][i - 1];
    ++C[A[i]][i];
  }
  m = max(C[1][n], C[2][n]);
  for (int i = 1; i < n; ++i) {
    m = max(m, C[1][i] + C[2][n] - C[2][i]);
  }
  for (int i = 1; i < n; ++i) {
    a = 0;
    for (int j = i - 1; j >= 0; --j) {
      c = C[1][j] + C[2][i] - C[2][j];
      a = max(a, c);
    }
    b = 0;
    for (int j = i + 1; j <= n; ++j) {
      c = C[1][j] - C[1][i] + C[2][n] - C[2][j];
      b = max(b, c);
    }
    m = max(m, a + b);
  }
  cout << m << endl;
  return 0;
}

```