#include <bits/stdc++.h>
using namespace std;
int n, a[2001], b[2001], d[2001], asc[2001], desc[2001], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[n - i - 1] = a[i];
  }
  for (int i = 0; i < n; i++) {
    d[i] = asc[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[i] < a[j]) continue;
      d[i] = max(d[i], d[j] + 1);
      asc[i] = max(asc[i - 1], d[i]);
    }
  }
  memset(d, 0, sizeof(d));
  for (int i = 0; i < n; i++) {
    d[i] = desc[i] = 1;
    for (int j = 0; j < i; j++) {
      if (b[i] > b[j]) continue;
      d[i] = max(d[i], d[j] + 1);
      desc[i] = max(desc[i - 1], d[i]);
    }
  }
  for (int i = 0; i < n; i++) ans = max(ans, asc[i] + desc[n - i - 2]);
  cout << ans;
}

### Prompt

Your task is to create 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;
int n, a[2001], b[2001], d[2001], asc[2001], desc[2001], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[n - i - 1] = a[i];
  }
  for (int i = 0; i < n; i++) {
    d[i] = asc[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[i] < a[j]) continue;
      d[i] = max(d[i], d[j] + 1);
      asc[i] = max(asc[i - 1], d[i]);
    }
  }
  memset(d, 0, sizeof(d));
  for (int i = 0; i < n; i++) {
    d[i] = desc[i] = 1;
    for (int j = 0; j < i; j++) {
      if (b[i] > b[j]) continue;
      d[i] = max(d[i], d[j] + 1);
      desc[i] = max(desc[i - 1], d[i]);
    }
  }
  for (int i = 0; i < n; i++) ans = max(ans, asc[i] + desc[n - i - 2]);
  cout << ans;
}

```