#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
int n, a[N], s1[N][N], s2[N][N], dp[5][N];
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      s1[i][j] = s1[i][j - 1] + (a[j] == 1);
      s2[i][j] = s2[i][j - 1] + (a[j] == 2);
    }
  for (int i = 1; i <= 4; ++i)
    for (int j = 1; j <= n; ++j)
      for (int k = 0; k <= j; ++k)
        if (i & 1)
          dp[i][j] = max(dp[i][j], dp[i - 1][k] + s1[k + 1][j]);
        else
          dp[i][j] = max(dp[i][j], dp[i - 1][k] + s2[k + 1][j]);
  printf("%d\n", dp[4][n]);
  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 N = 2005;
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
int n, a[N], s1[N][N], s2[N][N], dp[5][N];
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      s1[i][j] = s1[i][j - 1] + (a[j] == 1);
      s2[i][j] = s2[i][j - 1] + (a[j] == 2);
    }
  for (int i = 1; i <= 4; ++i)
    for (int j = 1; j <= n; ++j)
      for (int k = 0; k <= j; ++k)
        if (i & 1)
          dp[i][j] = max(dp[i][j], dp[i - 1][k] + s1[k + 1][j]);
        else
          dp[i][j] = max(dp[i][j], dp[i - 1][k] + s2[k + 1][j]);
  printf("%d\n", dp[4][n]);
  return 0;
}

```