#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 10;
int n;
int num[MAXN];
int L[MAXN][MAXN];
int R[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int ii = 1; ii <= n; ii++) {
    cin >> num[ii];
  }
  for (int ii = 1; ii <= n; ii++) {
    int dp1 = 0, dp2 = 0;
    for (int jj = ii; jj <= n; jj++) {
      if (num[jj] == 1)
        dp1++;
      else
        dp2 = max(dp1, dp2) + 1;
      L[ii][jj] = max(dp1, dp2);
    }
    dp1 = dp2 = 0;
    for (int jj = ii; jj <= n; jj++) {
      if (num[jj] == 2)
        dp2++;
      else
        dp1 = max(dp1, dp2) + 1;
      R[ii][jj] = max(dp1, dp2);
    }
  }
  int ans = 0;
  for (int ii = 1; ii <= n; ii++) {
    for (int jj = ii; jj <= n; jj++) {
      ans = max(ans, L[1][n] - L[ii][jj] + R[ii][jj]);
    }
  }
  cout << ans;
}

### Prompt

Please create a solution in cpp 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 MAXN = 2e3 + 10;
int n;
int num[MAXN];
int L[MAXN][MAXN];
int R[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int ii = 1; ii <= n; ii++) {
    cin >> num[ii];
  }
  for (int ii = 1; ii <= n; ii++) {
    int dp1 = 0, dp2 = 0;
    for (int jj = ii; jj <= n; jj++) {
      if (num[jj] == 1)
        dp1++;
      else
        dp2 = max(dp1, dp2) + 1;
      L[ii][jj] = max(dp1, dp2);
    }
    dp1 = dp2 = 0;
    for (int jj = ii; jj <= n; jj++) {
      if (num[jj] == 2)
        dp2++;
      else
        dp1 = max(dp1, dp2) + 1;
      R[ii][jj] = max(dp1, dp2);
    }
  }
  int ans = 0;
  for (int ii = 1; ii <= n; ii++) {
    for (int jj = ii; jj <= n; jj++) {
      ans = max(ans, L[1][n] - L[ii][jj] + R[ii][jj]);
    }
  }
  cout << ans;
}

```