#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  int arr[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int dp1[n];
  int dp2[n];
  int dp3[n];
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1) c++;
    dp2[i] = c;
  }
  c = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (arr[i] == 2) c++;
    dp3[i] = c;
  }
  for (int i = 0; i < n; i++) {
    int m = 0;
    for (int k = 0; k < i; k++) {
      int o = dp2[k], t = dp3[k + 1];
      if (i != n - 1) t -= dp3[i + 1];
      if (o && t) {
        m = max(m, o + t);
      }
    }
    dp1[i] = m;
  }
  int dp4[n][n];
  int dp6[n];
  c = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 2) c++;
    dp6[i] = c;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == j) {
        dp4[i][j] = 0;
      } else {
        if (arr[j] == 2)
          dp4[i][j] = dp4[i][j - 1];
        else {
          int x = dp6[j - 1];
          if (i) x -= dp6[i - 1];
          int y = dp4[i][j - 1];
          if (x && y)
            dp4[i][j] = max(x + 1, y + 1);
          else if (x && !y)
            dp4[i][j] = x + 1;
          else if (!x && y)
            dp4[i][j] = y + 1;
          else
            dp4[i][j] = 0;
        }
      }
    }
  }
  int mx = max(dp1[n - 1], dp2[n - 1]);
  mx = max(mx, dp3[0]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int co = dp4[i][j];
      if (co != 0) {
        if (i) co += dp2[i - 1];
        if (j != n - 1) co += dp3[j + 1];
        mx = max(mx, co);
      }
    }
  }
  cout << mx << endl;
  return 0;
}

### Prompt

Generate 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;
int main() {
  int n;
  scanf("%d", &n);
  int arr[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int dp1[n];
  int dp2[n];
  int dp3[n];
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1) c++;
    dp2[i] = c;
  }
  c = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (arr[i] == 2) c++;
    dp3[i] = c;
  }
  for (int i = 0; i < n; i++) {
    int m = 0;
    for (int k = 0; k < i; k++) {
      int o = dp2[k], t = dp3[k + 1];
      if (i != n - 1) t -= dp3[i + 1];
      if (o && t) {
        m = max(m, o + t);
      }
    }
    dp1[i] = m;
  }
  int dp4[n][n];
  int dp6[n];
  c = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 2) c++;
    dp6[i] = c;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == j) {
        dp4[i][j] = 0;
      } else {
        if (arr[j] == 2)
          dp4[i][j] = dp4[i][j - 1];
        else {
          int x = dp6[j - 1];
          if (i) x -= dp6[i - 1];
          int y = dp4[i][j - 1];
          if (x && y)
            dp4[i][j] = max(x + 1, y + 1);
          else if (x && !y)
            dp4[i][j] = x + 1;
          else if (!x && y)
            dp4[i][j] = y + 1;
          else
            dp4[i][j] = 0;
        }
      }
    }
  }
  int mx = max(dp1[n - 1], dp2[n - 1]);
  mx = max(mx, dp3[0]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int co = dp4[i][j];
      if (co != 0) {
        if (i) co += dp2[i - 1];
        if (j != n - 1) co += dp3[j + 1];
        mx = max(mx, co);
      }
    }
  }
  cout << mx << endl;
  return 0;
}

```