#include <bits/stdc++.h>
using namespace std;
long long int n;
long long int ar[2001];
long long int dp[2001][5];
long long int func(long long int ind, long long int j) {
  if (ind == n) return 0;
  if (j == 4) return 0;
  if (dp[ind][j] != -1) return dp[ind][j];
  long long int ret = func(ind + 1, j);
  ret = max(ret, func(ind, j + 1));
  if (j == 0 or j == 2) {
    if (ar[ind] == 1) {
      ret = max(ret, 1 + func(ind + 1, j + 1));
      ret = max(ret, 1 + func(ind + 1, j));
    }
  }
  if (j == 1 or j == 3) {
    if (ar[ind] == 2) {
      ret = max(ret, 1 + func(ind + 1, j + 1));
      ret = max(ret, 1 + func(ind + 1, j));
    }
  }
  return dp[ind][j] = ret;
}
int main() {
  long long int i, j, k, m, t;
  cin >> n;
  for (i = 0; i < n; i++) cin >> ar[i];
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= 4; j++) dp[i][j] = -1;
  }
  cout << func(0, 0);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long int n;
long long int ar[2001];
long long int dp[2001][5];
long long int func(long long int ind, long long int j) {
  if (ind == n) return 0;
  if (j == 4) return 0;
  if (dp[ind][j] != -1) return dp[ind][j];
  long long int ret = func(ind + 1, j);
  ret = max(ret, func(ind, j + 1));
  if (j == 0 or j == 2) {
    if (ar[ind] == 1) {
      ret = max(ret, 1 + func(ind + 1, j + 1));
      ret = max(ret, 1 + func(ind + 1, j));
    }
  }
  if (j == 1 or j == 3) {
    if (ar[ind] == 2) {
      ret = max(ret, 1 + func(ind + 1, j + 1));
      ret = max(ret, 1 + func(ind + 1, j));
    }
  }
  return dp[ind][j] = ret;
}
int main() {
  long long int i, j, k, m, t;
  cin >> n;
  for (i = 0; i < n; i++) cin >> ar[i];
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= 4; j++) dp[i][j] = -1;
  }
  cout << func(0, 0);
}

```