#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 17;
int dp[MAXN][MAXN][3];
int l[MAXN], r[MAXN];
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  l[0] = a[0] == 1;
  for (int i = 1; i < n; ++i) l[i] = l[i - 1] + (a[i] == 1);
  r[n - 1] = a[n - 1] == 2;
  for (int i = n - 2; i > -1; --i) r[i] = r[i + 1] + (a[i] == 2);
  reverse(a.begin(), a.end());
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1)
      dp[i][i][1] = 1;
    else
      dp[i][i][2] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (a[j] == 1) {
        dp[i][j][1] = dp[i][j - 1][1] + 1;
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = max(dp[i][j - 1][1] + 1, dp[i][j - 1][2] + 1);
      }
    }
  }
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      int z = (n - j - 1) > 0 ? l[(n - j - 1) - 1] : 0;
      int y = (n - i - 1) < n - 1 ? r[(n - i - 1) + 1] : 0;
      ans = max(ans, max(dp[i][j][1], dp[i][j][2]) + z + y);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP to 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;
const int MAXN = 2e3 + 17;
int dp[MAXN][MAXN][3];
int l[MAXN], r[MAXN];
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  l[0] = a[0] == 1;
  for (int i = 1; i < n; ++i) l[i] = l[i - 1] + (a[i] == 1);
  r[n - 1] = a[n - 1] == 2;
  for (int i = n - 2; i > -1; --i) r[i] = r[i + 1] + (a[i] == 2);
  reverse(a.begin(), a.end());
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1)
      dp[i][i][1] = 1;
    else
      dp[i][i][2] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (a[j] == 1) {
        dp[i][j][1] = dp[i][j - 1][1] + 1;
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = max(dp[i][j - 1][1] + 1, dp[i][j - 1][2] + 1);
      }
    }
  }
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      int z = (n - j - 1) > 0 ? l[(n - j - 1) - 1] : 0;
      int y = (n - i - 1) < n - 1 ? r[(n - i - 1) + 1] : 0;
      ans = max(ans, max(dp[i][j][1], dp[i][j][2]) + z + y);
    }
  }
  cout << ans << endl;
  return 0;
}

```