#include <bits/stdc++.h>
using namespace std;
void read(bool out = 0) {}
long long a[500009];
long long dp[500009];
long long get_one(long long x, long long y) {
  if (x > y) return 0;
  long long ans = dp[y];
  if (x) ans -= dp[x - 1];
  return ans;
}
long long get_two(long long x, long long y) {
  if (x > y) return 0;
  long long ans = dp[y];
  if (x) ans -= dp[x];
  ans = y - x + 1 - ans;
  return ans;
}
int main() {
  read();
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1)
    return cout << 1, 0;
  else if (n == 2)
    return cout << 2, 0;
  dp[0] = (a[0] == 1);
  for (int i = 1; i < n; i++) {
    dp[i] = (a[i] == 1) + dp[i - 1];
  }
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    long long test1 = 0;
    for (int j = 0; j <= i; j++) {
      test1 = max(test1, get_one(0, j - 1) + get_two(j, i));
      test1 = max(test1, get_one(0, j) + get_two(j + 1, i));
    }
    long long test2 = 0;
    for (int j = i + 1; j < n; j++) {
      test2 = max(test2, get_one(i + 1, j - 1) + get_two(j, n - 1));
      test2 = max(test2, get_one(i + 1, j) + get_two(j + 1, n - 1));
    }
    ans = max(ans, test1 + test2);
  }
  cout << ans << "\n";
}

### 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;
void read(bool out = 0) {}
long long a[500009];
long long dp[500009];
long long get_one(long long x, long long y) {
  if (x > y) return 0;
  long long ans = dp[y];
  if (x) ans -= dp[x - 1];
  return ans;
}
long long get_two(long long x, long long y) {
  if (x > y) return 0;
  long long ans = dp[y];
  if (x) ans -= dp[x];
  ans = y - x + 1 - ans;
  return ans;
}
int main() {
  read();
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1)
    return cout << 1, 0;
  else if (n == 2)
    return cout << 2, 0;
  dp[0] = (a[0] == 1);
  for (int i = 1; i < n; i++) {
    dp[i] = (a[i] == 1) + dp[i - 1];
  }
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    long long test1 = 0;
    for (int j = 0; j <= i; j++) {
      test1 = max(test1, get_one(0, j - 1) + get_two(j, i));
      test1 = max(test1, get_one(0, j) + get_two(j + 1, i));
    }
    long long test2 = 0;
    for (int j = i + 1; j < n; j++) {
      test2 = max(test2, get_one(i + 1, j - 1) + get_two(j, n - 1));
      test2 = max(test2, get_one(i + 1, j) + get_two(j + 1, n - 1));
    }
    ans = max(ans, test1 + test2);
  }
  cout << ans << "\n";
}

```