#include <bits/stdc++.h>
int main(int argc, char** argv) {
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(0);
  std::cout << std::fixed << std::setprecision(6);
  int k, a[2222], acc1[2222], acc2[2222];
  std::cin >> k;
  acc1[0] = acc2[0] = 0;
  for (int i = 0; i < k; i++) std::cin >> a[i];
  std::reverse(a, a + k);
  for (int i = 0; i < k; i++) {
    acc1[i + 1] = acc1[i] + (a[i] == 1 ? 1 : 0);
    acc2[i + 1] = acc2[i] + (a[i] == 2 ? 1 : 0);
  }
  int cp = 0;
  for (int i = 0; i < k; i++) {
    int dp1 = 0, dp2 = 0;
    for (int j = i; j < k; j++) {
      if (a[j] == 1) {
        dp1++;
      } else {
        dp2 = std::max(dp1 + 1, dp2 + 1);
      }
      cp = std::max(acc1[k] - acc1[j + 1] + std::max(dp1, dp2) + acc2[i], cp);
    }
  }
  std::cout << cp << std::endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution 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>
int main(int argc, char** argv) {
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(0);
  std::cout << std::fixed << std::setprecision(6);
  int k, a[2222], acc1[2222], acc2[2222];
  std::cin >> k;
  acc1[0] = acc2[0] = 0;
  for (int i = 0; i < k; i++) std::cin >> a[i];
  std::reverse(a, a + k);
  for (int i = 0; i < k; i++) {
    acc1[i + 1] = acc1[i] + (a[i] == 1 ? 1 : 0);
    acc2[i + 1] = acc2[i] + (a[i] == 2 ? 1 : 0);
  }
  int cp = 0;
  for (int i = 0; i < k; i++) {
    int dp1 = 0, dp2 = 0;
    for (int j = i; j < k; j++) {
      if (a[j] == 1) {
        dp1++;
      } else {
        dp2 = std::max(dp1 + 1, dp2 + 1);
      }
      cp = std::max(acc1[k] - acc1[j + 1] + std::max(dp1, dp2) + acc2[i], cp);
    }
  }
  std::cout << cp << std::endl;
  return 0;
}

```