#include <bits/stdc++.h>
using namespace std;
inline void Rd(int &res) {
  char c;
  res = 0;
  while (c = getchar(), c < '0')
    ;
  do {
    res = (res << 1) + (res << 3) + (c ^ 48);
  } while (c = getchar(), c >= '0');
}
const int M = 2005;
int n, A[M], sum[3][M], suf[M];
int main() {
  Rd(n);
  for (int i = 1; i <= n; i++) Rd(A[i]);
  for (int i = 1; i <= n; i++) {
    sum[1][i] = sum[1][i - 1];
    sum[2][i] = sum[2][i - 1];
    sum[A[i]][i]++;
  }
  int ans = 0;
  for (int i = 0; i <= n; i++)
    ans = max(ans, sum[1][i] + sum[2][n] - sum[2][i]);
  for (int i = 1; i <= n; i++)
    for (int j = i - 1; j <= n; j++)
      suf[i] = max(suf[i], sum[1][j] - sum[1][i - 1] + sum[2][n] - sum[2][j]);
  for (int i = 1; i <= n; i++) {
    int pre12 = 0;
    for (int j = 0; j < i; j++)
      pre12 = max(pre12, sum[1][j] + sum[2][i - 1] - sum[2][j]);
    int tmp = sum[2][i - 1] - sum[1][i - 1];
    for (int j = i; j <= n; j++) {
      int l1 = sum[1][j] - sum[1][i - 1];
      int l2 = sum[2][j] - sum[2][i - 1];
      tmp = max(tmp, sum[2][j] - sum[1][j]);
      int l21 = sum[1][j] - sum[2][i - 1] + tmp;
      ans = max(ans, pre12 + l2 + sum[2][n] - sum[2][j]);
      ans = max(ans, sum[1][i - 1] + l21 + sum[2][n] - sum[2][j]);
      ans = max(ans, sum[1][i - 1] + l1 + suf[j + 1]);
    }
  }
  printf("%d\n", 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;
inline void Rd(int &res) {
  char c;
  res = 0;
  while (c = getchar(), c < '0')
    ;
  do {
    res = (res << 1) + (res << 3) + (c ^ 48);
  } while (c = getchar(), c >= '0');
}
const int M = 2005;
int n, A[M], sum[3][M], suf[M];
int main() {
  Rd(n);
  for (int i = 1; i <= n; i++) Rd(A[i]);
  for (int i = 1; i <= n; i++) {
    sum[1][i] = sum[1][i - 1];
    sum[2][i] = sum[2][i - 1];
    sum[A[i]][i]++;
  }
  int ans = 0;
  for (int i = 0; i <= n; i++)
    ans = max(ans, sum[1][i] + sum[2][n] - sum[2][i]);
  for (int i = 1; i <= n; i++)
    for (int j = i - 1; j <= n; j++)
      suf[i] = max(suf[i], sum[1][j] - sum[1][i - 1] + sum[2][n] - sum[2][j]);
  for (int i = 1; i <= n; i++) {
    int pre12 = 0;
    for (int j = 0; j < i; j++)
      pre12 = max(pre12, sum[1][j] + sum[2][i - 1] - sum[2][j]);
    int tmp = sum[2][i - 1] - sum[1][i - 1];
    for (int j = i; j <= n; j++) {
      int l1 = sum[1][j] - sum[1][i - 1];
      int l2 = sum[2][j] - sum[2][i - 1];
      tmp = max(tmp, sum[2][j] - sum[1][j]);
      int l21 = sum[1][j] - sum[2][i - 1] + tmp;
      ans = max(ans, pre12 + l2 + sum[2][n] - sum[2][j]);
      ans = max(ans, sum[1][i - 1] + l21 + sum[2][n] - sum[2][j]);
      ans = max(ans, sum[1][i - 1] + l1 + suf[j + 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```