#include <bits/stdc++.h>
using namespace std;
int c[3055], a1[3055], a2[3055];
int dp[3055][3055][3];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n; i++) {
    a1[i] += a1[i - 1] + (c[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    a2[i] += a2[i + 1] + (c[i] == 2);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      dp[i][j][2] = dp[i][j - 1][2];
      dp[i][j][1] = dp[i][j - 1][1];
      if (c[j] == 2) dp[i][j][2] = dp[i][j - 1][2] + 1;
      if (c[j] == 1) dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(ans, a1[i - 1] + a2[j + 1] + dp[i][j][1]);
      ans = max(ans, a1[i - 1] + a2[j + 1] + dp[i][j][2]);
    }
  }
  cout << 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 c[3055], a1[3055], a2[3055];
int dp[3055][3055][3];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n; i++) {
    a1[i] += a1[i - 1] + (c[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    a2[i] += a2[i + 1] + (c[i] == 2);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      dp[i][j][2] = dp[i][j - 1][2];
      dp[i][j][1] = dp[i][j - 1][1];
      if (c[j] == 2) dp[i][j][2] = dp[i][j - 1][2] + 1;
      if (c[j] == 1) dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + 1;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(ans, a1[i - 1] + a2[j + 1] + dp[i][j][1]);
      ans = max(ans, a1[i - 1] + a2[j + 1] + dp[i][j][2]);
    }
  }
  cout << ans;
  return 0;
}

```