#include <bits/stdc++.h>
using namespace std;
inline int R() {
  char c;
  int sign = 1, res = 0;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') sign = -1;
  res += c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  return res * sign;
}
const int Maxn = 2e3 + 5;
int a[Maxn], n, s[Maxn], dp[Maxn][Maxn];
int main() {
  n = R();
  for (int i = 1; i <= n; i++) {
    a[i] = R() - 1;
    s[i] = s[i - 1] + a[i];
  }
  for (int i = 1; i <= n; i++) dp[i][i] = 1;
  for (int k = 1; k <= n; k++)
    for (int i = 1; i + k <= n; i++) {
      int j = k + i;
      if (a[i] == 0)
        dp[i][j] = max(dp[i + 1][j],
                       max(s[j] - s[i - 1], j - i + 1 - (s[j] - s[i - 1])));
      if (a[i] == 1)
        dp[i][j] = max(dp[i + 1][j] + 1,
                       max(s[j] - s[i - 1], j - i + 1 - (s[j] - s[i - 1])));
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      ans = max(dp[i][j] + i - 1 - s[i - 1] + s[n] - s[j], ans);
  printf("%d\n", ans);
}

### 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 int R() {
  char c;
  int sign = 1, res = 0;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') sign = -1;
  res += c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  return res * sign;
}
const int Maxn = 2e3 + 5;
int a[Maxn], n, s[Maxn], dp[Maxn][Maxn];
int main() {
  n = R();
  for (int i = 1; i <= n; i++) {
    a[i] = R() - 1;
    s[i] = s[i - 1] + a[i];
  }
  for (int i = 1; i <= n; i++) dp[i][i] = 1;
  for (int k = 1; k <= n; k++)
    for (int i = 1; i + k <= n; i++) {
      int j = k + i;
      if (a[i] == 0)
        dp[i][j] = max(dp[i + 1][j],
                       max(s[j] - s[i - 1], j - i + 1 - (s[j] - s[i - 1])));
      if (a[i] == 1)
        dp[i][j] = max(dp[i + 1][j] + 1,
                       max(s[j] - s[i - 1], j - i + 1 - (s[j] - s[i - 1])));
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      ans = max(dp[i][j] + i - 1 - s[i - 1] + s[n] - s[j], ans);
  printf("%d\n", ans);
}

```