#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000 + 10;
int sum[maxn][3], a[maxn], n, f[maxn][3], ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sum[i][1] = sum[i - 1][1];
    sum[i][2] = sum[i - 1][2];
    sum[i][a[i]]++;
  }
  for (int l = 1; l <= n; l++) {
    f[l - 1][1] = f[l - 1][2] = 0;
    for (int r = l; r <= n; r++) {
      f[r][1] = f[r - 1][1];
      f[r][2] = f[r - 1][2];
      if (a[r] == 2)
        f[r][2]++;
      else
        f[r][1]++;
      f[r][1] = max(f[r][1], f[r][2]);
      int len = max(f[r][1], f[r][2]);
      ans = max(ans, sum[l - 1][1] + len + sum[n][2] - sum[r][2]);
    }
  }
  f[0][1] = f[0][2] = 0;
  for (int i = 1; i <= n; i++) {
    f[i][1] = f[i - 1][1];
    f[i][2] = f[i - 1][2];
    if (a[i] == 1)
      f[i][1]++;
    else
      f[i][2]++;
    f[i][2] = max(f[i][2], f[i][1]);
  }
  ans = max(ans, max(f[n][1], f[n][2]));
  printf("%d\n", ans);
}

### 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>
using namespace std;
const int maxn = 2000 + 10;
int sum[maxn][3], a[maxn], n, f[maxn][3], ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sum[i][1] = sum[i - 1][1];
    sum[i][2] = sum[i - 1][2];
    sum[i][a[i]]++;
  }
  for (int l = 1; l <= n; l++) {
    f[l - 1][1] = f[l - 1][2] = 0;
    for (int r = l; r <= n; r++) {
      f[r][1] = f[r - 1][1];
      f[r][2] = f[r - 1][2];
      if (a[r] == 2)
        f[r][2]++;
      else
        f[r][1]++;
      f[r][1] = max(f[r][1], f[r][2]);
      int len = max(f[r][1], f[r][2]);
      ans = max(ans, sum[l - 1][1] + len + sum[n][2] - sum[r][2]);
    }
  }
  f[0][1] = f[0][2] = 0;
  for (int i = 1; i <= n; i++) {
    f[i][1] = f[i - 1][1];
    f[i][2] = f[i - 1][2];
    if (a[i] == 1)
      f[i][1]++;
    else
      f[i][2]++;
    f[i][2] = max(f[i][2], f[i][1]);
  }
  ans = max(ans, max(f[n][1], f[n][2]));
  printf("%d\n", ans);
}

```