#include <bits/stdc++.h>
using namespace std;
int a[3000];
int dp[3000][5];
int lis(int i, int j, int s) {
  if (i > j) return 0;
  if (dp[i][s] != -1) return dp[i][s];
  if (s == 0) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 0) + 1, lis(i + 1, j, 0));
    else
      return dp[i][s] = max(lis(i + 1, j, 1) + 1, lis(i + 1, j, 0));
  } else if (s == 1) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 2) + 1, lis(i + 1, j, 1));
    else
      return dp[i][s] = max(lis(i + 1, j, 1) + 1, lis(i + 1, j, 1));
  } else if (s == 2) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 2) + 1, lis(i + 1, j, 2));
    else
      return dp[i][s] = max(lis(i + 1, j, 3) + 1, lis(i + 1, j, 2));
  } else {
    if (a[i] == 1)
      return dp[i][s] = lis(i + 1, j, 3);
    else
      return dp[i][s] = lis(i + 1, j, 3) + 1;
  }
}
int main() {
  int n;
  cin >> n;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; ++i) cin >> a[i];
  cout << lis(0, n - 1, 0) << endl;
}

### Prompt

In Cpp, your task is to solve 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[3000];
int dp[3000][5];
int lis(int i, int j, int s) {
  if (i > j) return 0;
  if (dp[i][s] != -1) return dp[i][s];
  if (s == 0) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 0) + 1, lis(i + 1, j, 0));
    else
      return dp[i][s] = max(lis(i + 1, j, 1) + 1, lis(i + 1, j, 0));
  } else if (s == 1) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 2) + 1, lis(i + 1, j, 1));
    else
      return dp[i][s] = max(lis(i + 1, j, 1) + 1, lis(i + 1, j, 1));
  } else if (s == 2) {
    if (a[i] == 1)
      return dp[i][s] = max(lis(i + 1, j, 2) + 1, lis(i + 1, j, 2));
    else
      return dp[i][s] = max(lis(i + 1, j, 3) + 1, lis(i + 1, j, 2));
  } else {
    if (a[i] == 1)
      return dp[i][s] = lis(i + 1, j, 3);
    else
      return dp[i][s] = lis(i + 1, j, 3) + 1;
  }
}
int main() {
  int n;
  cin >> n;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; ++i) cin >> a[i];
  cout << lis(0, n - 1, 0) << endl;
}

```