#include <bits/stdc++.h>
using namespace std;
int S[2002], n, t;
int C[2002][2002];
int ML[2002][2002], MR[2002][2002];
int ans = -10000, M;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> t;
    if (t == 2)
      S[i] = S[i - 1] + 1;
    else
      S[i] = S[i - 1];
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      ML[i][j] = -10000;
      MR[i][j] = 10000;
    }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) C[i][j] = S[j] - S[i - 1];
  for (int i = 0; i <= n - 1; i++)
    for (int j = i + 1; j <= n; j++) {
      ML[j][i] = max(ML[j - 1][i], C[j - i][j]);
    }
  for (int i = 0; i <= n - 1; i++)
    for (int j = n - i; j >= 1; j--) {
      MR[j][i] = min(MR[j + 1][i], C[j][i + j]);
    }
  for (int i = 1; i <= n + 1; i++) {
    M = 0;
    for (int j = 1; (j <= n - i + 1) && (j <= i - 1); j++) {
      M = max(M, ML[i - 1][j - 1] - C[i][i + j - 1]);
    }
    for (int j = 1; j <= i - 1 && j <= n - i - 1; j++) {
      M = max(M, C[i - j][i - 1] - MR[i][j - 1]);
    }
    ans = max(ans, i - 1 + 2 * M + S[n] - 2 * S[i - 1]);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 S[2002], n, t;
int C[2002][2002];
int ML[2002][2002], MR[2002][2002];
int ans = -10000, M;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> t;
    if (t == 2)
      S[i] = S[i - 1] + 1;
    else
      S[i] = S[i - 1];
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      ML[i][j] = -10000;
      MR[i][j] = 10000;
    }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) C[i][j] = S[j] - S[i - 1];
  for (int i = 0; i <= n - 1; i++)
    for (int j = i + 1; j <= n; j++) {
      ML[j][i] = max(ML[j - 1][i], C[j - i][j]);
    }
  for (int i = 0; i <= n - 1; i++)
    for (int j = n - i; j >= 1; j--) {
      MR[j][i] = min(MR[j + 1][i], C[j][i + j]);
    }
  for (int i = 1; i <= n + 1; i++) {
    M = 0;
    for (int j = 1; (j <= n - i + 1) && (j <= i - 1); j++) {
      M = max(M, ML[i - 1][j - 1] - C[i][i + j - 1]);
    }
    for (int j = 1; j <= i - 1 && j <= n - i - 1; j++) {
      M = max(M, C[i - j][i - 1] - MR[i][j - 1]);
    }
    ans = max(ans, i - 1 + 2 * M + S[n] - 2 * S[i - 1]);
  }
  cout << ans << endl;
  return 0;
}

```