#include <bits/stdc++.h>
using namespace std;
const int DIM = 2005;
int cnt[DIM][3], arr[DIM], dpl[DIM][DIM][3], dpr[DIM][DIM][3];
int f(int l, int r, int c) { return cnt[r][c] - cnt[l - 1][c]; }
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    cnt[i][1] = cnt[i - 1][1];
    cnt[i][2] = cnt[i - 1][2];
    cnt[i][arr[i]]++;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      dpl[i][j][1] = dpl[i][j - 1][1];
      dpl[i][j][2] = max(dpl[i][j - 1][1], dpl[i][j - 1][2]);
      dpl[i][j][arr[j]]++;
      dpl[i][j][2] = max(dpl[i][j][2], dpl[i][j][1]);
    }
    for (int j = i; j >= 1; --j) {
      dpr[i][j][1] = dpr[i][j + 1][1];
      dpr[i][j][2] = max(dpr[i][j + 1][1], dpr[i][j + 1][2]);
      dpr[i][j][arr[j]]++;
      dpr[i][j][2] = max(dpr[i][j][2], dpr[i][j][1]);
    }
  }
  int ans = dpl[1][n][2];
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      ans = max(max(ans, dpl[1][i - 1][2] + f(i, n, 2)),
                max(f(1, i - 1, 1) + dpr[j][i][2] + f(j + 1, n, 2),
                    f(1, j, 1) + dpl[j + 1][n][2]));
    }
  }
  cout << ans;
  return 0;
}

### 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;
const int DIM = 2005;
int cnt[DIM][3], arr[DIM], dpl[DIM][DIM][3], dpr[DIM][DIM][3];
int f(int l, int r, int c) { return cnt[r][c] - cnt[l - 1][c]; }
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    cnt[i][1] = cnt[i - 1][1];
    cnt[i][2] = cnt[i - 1][2];
    cnt[i][arr[i]]++;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      dpl[i][j][1] = dpl[i][j - 1][1];
      dpl[i][j][2] = max(dpl[i][j - 1][1], dpl[i][j - 1][2]);
      dpl[i][j][arr[j]]++;
      dpl[i][j][2] = max(dpl[i][j][2], dpl[i][j][1]);
    }
    for (int j = i; j >= 1; --j) {
      dpr[i][j][1] = dpr[i][j + 1][1];
      dpr[i][j][2] = max(dpr[i][j + 1][1], dpr[i][j + 1][2]);
      dpr[i][j][arr[j]]++;
      dpr[i][j][2] = max(dpr[i][j][2], dpr[i][j][1]);
    }
  }
  int ans = dpl[1][n][2];
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      ans = max(max(ans, dpl[1][i - 1][2] + f(i, n, 2)),
                max(f(1, i - 1, 1) + dpr[j][i][2] + f(j + 1, n, 2),
                    f(1, j, 1) + dpl[j + 1][n][2]));
    }
  }
  cout << ans;
  return 0;
}

```