#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
int f[2010][2010], n, a[2010], sum1[2010], sum2[2010];
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    sum1[i] = sum1[i - 1] + (a[i] == 1);
    sum2[i] = sum2[i - 1] + (a[i] == 2);
  }
  for (register int i = 1; i <= n; ++i) {
    register int now = i;
    f[i][i] = 1;
    for (register int j = i + 1; j <= n; ++j) {
      register int tmp;
      for (register int k = now; k <= j; ++k) {
        if (sum2[k] - sum2[i - 1] + sum1[j] - sum1[k] > f[i][j]) {
          f[i][j] = sum2[k] - sum2[i - 1] + sum1[j] - sum1[k];
          tmp = k;
        }
      }
      now = tmp;
    }
  }
  register int ans = 0;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = i; j <= n; ++j) {
      ans = std::max(ans, sum1[i - 1] - sum1[0] + sum2[n] - sum2[j] + f[i][j]);
    }
  }
  printf("%d\n", ans);
}

### 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>
#pragma GCC optimize(3)
using namespace std;
int f[2010][2010], n, a[2010], sum1[2010], sum2[2010];
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    sum1[i] = sum1[i - 1] + (a[i] == 1);
    sum2[i] = sum2[i - 1] + (a[i] == 2);
  }
  for (register int i = 1; i <= n; ++i) {
    register int now = i;
    f[i][i] = 1;
    for (register int j = i + 1; j <= n; ++j) {
      register int tmp;
      for (register int k = now; k <= j; ++k) {
        if (sum2[k] - sum2[i - 1] + sum1[j] - sum1[k] > f[i][j]) {
          f[i][j] = sum2[k] - sum2[i - 1] + sum1[j] - sum1[k];
          tmp = k;
        }
      }
      now = tmp;
    }
  }
  register int ans = 0;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = i; j <= n; ++j) {
      ans = std::max(ans, sum1[i - 1] - sum1[0] + sum2[n] - sum2[j] + f[i][j]);
    }
  }
  printf("%d\n", ans);
}

```