#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  int N;
  cin >> N;
  vi A(N);
  for (int(i) = (0); (i) < (int)(N); ++(i)) cin >> A[i];
  static int dp[2001][2][2];
  memset((dp), (-1), sizeof(dp));
  dp[0][0][0] = 0;
  static tuple<int, int, int> g[2001][2001];
  for (int r = N; r >= 1; --r) {
    int o = 0, ot = 0, t = 0;
    for (int i = r - 1; i >= 0; --i) {
      ot = max(o, ot) + (A[i] == 2);
      if (A[i] == 1)
        o++;
      else
        t++;
      g[i][r] = make_tuple(o, ot, t);
    }
  }
  for (int(i) = (0); (i) < (int)(N); ++(i))
    for (int(j) = (0); (j) < (int)(2); ++(j))
      for (int(k) = (0); (k) < (int)(2); ++(k)) {
        int &cu = dp[i][j][k];
        if (cu == -1) continue;
        if (!j) {
          for (int r = i + 1; r <= N; ++r) {
            int o, ot, t;
            tie(o, ot, t) = g[i][r];
            auto &ne = dp[r][1];
            if (!k) {
              (ne[0]) = max((ne[0]), (cu + o));
              (ne[1]) = max((ne[1]), (cu + ot));
            } else {
              (ne[1]) = max((ne[1]), (cu + t));
            }
          }
        }
        auto &ne = dp[i + 1][j];
        if (A[i] == 2) (ne[1]) = max((ne[1]), (cu + 1));
        if (!k && A[i] == 1) {
          (ne[0]) = max((ne[0]), (cu + 1));
        }
        (ne[k]) = max((ne[k]), (cu));
      }
  int ans = 0;
  for (int(i) = (0); (i) < (int)(2); ++(i)) (ans) = max((ans), (dp[N][1][i]));
  cout << ans << endl;
}

### 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;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  int N;
  cin >> N;
  vi A(N);
  for (int(i) = (0); (i) < (int)(N); ++(i)) cin >> A[i];
  static int dp[2001][2][2];
  memset((dp), (-1), sizeof(dp));
  dp[0][0][0] = 0;
  static tuple<int, int, int> g[2001][2001];
  for (int r = N; r >= 1; --r) {
    int o = 0, ot = 0, t = 0;
    for (int i = r - 1; i >= 0; --i) {
      ot = max(o, ot) + (A[i] == 2);
      if (A[i] == 1)
        o++;
      else
        t++;
      g[i][r] = make_tuple(o, ot, t);
    }
  }
  for (int(i) = (0); (i) < (int)(N); ++(i))
    for (int(j) = (0); (j) < (int)(2); ++(j))
      for (int(k) = (0); (k) < (int)(2); ++(k)) {
        int &cu = dp[i][j][k];
        if (cu == -1) continue;
        if (!j) {
          for (int r = i + 1; r <= N; ++r) {
            int o, ot, t;
            tie(o, ot, t) = g[i][r];
            auto &ne = dp[r][1];
            if (!k) {
              (ne[0]) = max((ne[0]), (cu + o));
              (ne[1]) = max((ne[1]), (cu + ot));
            } else {
              (ne[1]) = max((ne[1]), (cu + t));
            }
          }
        }
        auto &ne = dp[i + 1][j];
        if (A[i] == 2) (ne[1]) = max((ne[1]), (cu + 1));
        if (!k && A[i] == 1) {
          (ne[0]) = max((ne[0]), (cu + 1));
        }
        (ne[k]) = max((ne[k]), (cu));
      }
  int ans = 0;
  for (int(i) = (0); (i) < (int)(2); ++(i)) (ans) = max((ans), (dp[N][1][i]));
  cout << ans << endl;
}

```