#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 7;
int sum[N][2];
vector<int> vec, w;
int main() {
  int n;
  int cnt = 0, flag = 0, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    if (a != flag) {
      vec.push_back(flag);
      w.push_back(cnt);
      cnt = 1;
      flag = a;
    } else {
      cnt++;
    }
  }
  vec.push_back(flag);
  w.push_back(cnt);
  int m = vec.size() - 1;
  for (int i = 1; i <= m; i++) {
    sum[i][0] = sum[i - 1][0] + (vec[i] == 1 ? w[i] : 0);
    sum[i][1] = sum[i - 1][1] + (vec[i] == 2 ? w[i] : 0);
  }
  for (int i = 0; i <= m; i++) {
    ans = max(sum[i][0] + sum[m][1] - sum[i][1], ans);
  }
  for (int i = 1; i < m; i++) {
    for (int j = i; j <= m; j++) {
      if (vec[i] == 2 && vec[j] == 1) {
        int cnt = sum[i][0] + sum[m][1] - sum[j][1];
        for (int k = i; k <= j; k++) {
          ans =
              max(ans, cnt + sum[k][1] - sum[i - 1][1] + sum[j][0] - sum[k][0]);
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int N = 2e3 + 7;
int sum[N][2];
vector<int> vec, w;
int main() {
  int n;
  int cnt = 0, flag = 0, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    if (a != flag) {
      vec.push_back(flag);
      w.push_back(cnt);
      cnt = 1;
      flag = a;
    } else {
      cnt++;
    }
  }
  vec.push_back(flag);
  w.push_back(cnt);
  int m = vec.size() - 1;
  for (int i = 1; i <= m; i++) {
    sum[i][0] = sum[i - 1][0] + (vec[i] == 1 ? w[i] : 0);
    sum[i][1] = sum[i - 1][1] + (vec[i] == 2 ? w[i] : 0);
  }
  for (int i = 0; i <= m; i++) {
    ans = max(sum[i][0] + sum[m][1] - sum[i][1], ans);
  }
  for (int i = 1; i < m; i++) {
    for (int j = i; j <= m; j++) {
      if (vec[i] == 2 && vec[j] == 1) {
        int cnt = sum[i][0] + sum[m][1] - sum[j][1];
        for (int k = i; k <= j; k++) {
          ans =
              max(ans, cnt + sum[k][1] - sum[i - 1][1] + sum[j][0] - sum[k][0]);
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```