#include <bits/stdc++.h>
using namespace std;
int dp[2004][2004];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> v(n + 2);
  vector<int> one(n + 2);
  vector<int> two(n + 2);
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 1; i <= n; i++) {
    one[i] += one[i - 1] + (v[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    two[i] += two[i + 1] + (v[i] == 2);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      dp[i][j] = max({dp[i][j], one[j] - one[i - 1], two[i] - two[j + 1]});
    }
  }
  int ans = 1;
  for (int len = 1; len < n; len++) {
    for (int start = 1; start <= n - len; start++) {
      dp[start][start + len] =
          max({dp[start][start + len], dp[start + 1][start + len - 1] +
                                           (v[start] == 2) +
                                           (v[start + len] == 1)});
    }
  }
  for (int start = 1; start <= n; start++) {
    for (int end = start; end <= n; end++) {
      ans = max(ans, one[start - 1] + dp[start][end] + two[end + 1]);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 dp[2004][2004];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> v(n + 2);
  vector<int> one(n + 2);
  vector<int> two(n + 2);
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 1; i <= n; i++) {
    one[i] += one[i - 1] + (v[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    two[i] += two[i + 1] + (v[i] == 2);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      dp[i][j] = max({dp[i][j], one[j] - one[i - 1], two[i] - two[j + 1]});
    }
  }
  int ans = 1;
  for (int len = 1; len < n; len++) {
    for (int start = 1; start <= n - len; start++) {
      dp[start][start + len] =
          max({dp[start][start + len], dp[start + 1][start + len - 1] +
                                           (v[start] == 2) +
                                           (v[start + len] == 1)});
    }
  }
  for (int start = 1; start <= n; start++) {
    for (int end = start; end <= n; end++) {
      ans = max(ans, one[start - 1] + dp[start][end] + two[end + 1]);
    }
  }
  cout << ans << '\n';
  return 0;
}

```