#include <bits/stdc++.h>
using namespace std;
int a[2005], pre[2005], lst[2005], dp[2005][2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    pre[i] = pre[i - 1];
    if (a[i] == 1) ++pre[i];
  }
  for (int i = n; i; --i) {
    lst[i] = lst[i + 1];
    if (a[i] == 2) ++lst[i];
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    memset(dp, 0, sizeof(dp));
    for (int j = i + 1; j <= n; ++j) {
      dp[j][0] = dp[j - 1][0];
      dp[j][1] = dp[j - 1][1];
      if (a[j] == 1)
        dp[j][0] = max(dp[j][0] + 1, dp[j - 1][1] + 1);
      else
        ++dp[j][1];
    }
    for (int j = i + 1; j <= n + 1; ++j) {
      ans = max(ans, pre[i] + lst[j] + max(dp[j - 1][0], dp[j - 1][1]));
    }
  }
  printf("%d\n", ans);
  return 0;
}

### 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;
int a[2005], pre[2005], lst[2005], dp[2005][2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    pre[i] = pre[i - 1];
    if (a[i] == 1) ++pre[i];
  }
  for (int i = n; i; --i) {
    lst[i] = lst[i + 1];
    if (a[i] == 2) ++lst[i];
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    memset(dp, 0, sizeof(dp));
    for (int j = i + 1; j <= n; ++j) {
      dp[j][0] = dp[j - 1][0];
      dp[j][1] = dp[j - 1][1];
      if (a[j] == 1)
        dp[j][0] = max(dp[j][0] + 1, dp[j - 1][1] + 1);
      else
        ++dp[j][1];
    }
    for (int j = i + 1; j <= n + 1; ++j) {
      ans = max(ans, pre[i] + lst[j] + max(dp[j - 1][0], dp[j - 1][1]));
    }
  }
  printf("%d\n", ans);
  return 0;
}

```