#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
short dp[2020][2020][3][3], dp1[2020][2020][3][3];
int n, a[2020], b[2020], ans = -1;
void calc_up() {
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      if (a[r] == 1) {
        dp[l][r][1][1] = dp[l][r - 1][1][1] + 1;
        dp[l][r][1][2] = dp[l][r - 1][1][2];
        dp[l][r][2][2] = dp[l][r - 1][2][2];
      } else {
        dp[l][r][1][1] = dp[l][r - 1][1][1];
        dp[l][r][1][2] = max(dp[l][r - 1][1][2], dp[l][r - 1][1][1]) + 1;
        dp[l][r][2][2] = dp[l][r - 1][2][2] + 1;
      }
    }
  }
}
void calc_down() {
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      if (a[r] == 1) {
        dp1[l][r][1][1] = dp1[l][r - 1][1][1] + 1;
        dp1[l][r][2][1] = max(dp1[l][r - 1][2][1], dp1[l][r - 1][2][2]) + 1;
        dp1[l][r][2][2] = dp1[l][r - 1][2][2];
      } else {
        dp1[l][r][1][1] = dp1[l][r - 1][1][1];
        dp1[l][r][2][1] = dp1[l][r - 1][2][1];
        dp1[l][r][2][2] = dp1[l][r - 1][2][2] + 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  calc_up();
  calc_down();
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int cur = max(dp[1][l - 1][1][2], dp[1][l - 1][2][2]) + dp[l][r][2][2];
      int cur1 = dp[1][l - 1][1][1] + dp1[l][r][1][1] +
                 max(dp[r + 1][n][1][2], dp[r + 1][n][2][2]);
      int cur2 = dp[1][l - 1][1][1] + dp1[l][r][2][1] + dp[r + 1][n][2][2];
      int cur3 = dp[1][l - 1][1][1] + dp1[l][r][2][2] + dp[r + 1][n][2][2];
      ans = max(ans, max(max(cur1, cur2), max(cur, cur3)));
    }
  }
  cout << ans << endl;
}

### 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;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
short dp[2020][2020][3][3], dp1[2020][2020][3][3];
int n, a[2020], b[2020], ans = -1;
void calc_up() {
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      if (a[r] == 1) {
        dp[l][r][1][1] = dp[l][r - 1][1][1] + 1;
        dp[l][r][1][2] = dp[l][r - 1][1][2];
        dp[l][r][2][2] = dp[l][r - 1][2][2];
      } else {
        dp[l][r][1][1] = dp[l][r - 1][1][1];
        dp[l][r][1][2] = max(dp[l][r - 1][1][2], dp[l][r - 1][1][1]) + 1;
        dp[l][r][2][2] = dp[l][r - 1][2][2] + 1;
      }
    }
  }
}
void calc_down() {
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      if (a[r] == 1) {
        dp1[l][r][1][1] = dp1[l][r - 1][1][1] + 1;
        dp1[l][r][2][1] = max(dp1[l][r - 1][2][1], dp1[l][r - 1][2][2]) + 1;
        dp1[l][r][2][2] = dp1[l][r - 1][2][2];
      } else {
        dp1[l][r][1][1] = dp1[l][r - 1][1][1];
        dp1[l][r][2][1] = dp1[l][r - 1][2][1];
        dp1[l][r][2][2] = dp1[l][r - 1][2][2] + 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  calc_up();
  calc_down();
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int cur = max(dp[1][l - 1][1][2], dp[1][l - 1][2][2]) + dp[l][r][2][2];
      int cur1 = dp[1][l - 1][1][1] + dp1[l][r][1][1] +
                 max(dp[r + 1][n][1][2], dp[r + 1][n][2][2]);
      int cur2 = dp[1][l - 1][1][1] + dp1[l][r][2][1] + dp[r + 1][n][2][2];
      int cur3 = dp[1][l - 1][1][1] + dp1[l][r][2][2] + dp[r + 1][n][2][2];
      ans = max(ans, max(max(cur1, cur2), max(cur, cur3)));
    }
  }
  cout << ans << endl;
}

```