#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2005;
long long int n, a[MAXN], dp1[MAXN], dp2[MAXN], oc[MAXN], tc[MAXN], res;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  if (a[0] == 1)
    oc[0] = 1;
  else
    dp1[0] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == 1) {
      oc[i] = oc[i - 1] + 1;
      dp1[i] = dp1[i - 1];
    } else {
      dp1[i] = max(oc[i - 1], dp1[i - 1]) + 1;
      oc[i] = oc[i - 1];
    }
  }
  if (a[n - 1] == 2)
    tc[n - 1] = 1;
  else
    dp2[n - 1] = 1;
  for (int i = n - 2; i >= 0; i--) {
    if (a[i] == 1) {
      tc[i] = tc[i + 1];
      dp2[i] = max(tc[i + 1], dp2[i + 1]) + 1;
    } else {
      tc[i] = tc[i + 1] + 1;
      dp2[i] = dp2[i + 1];
    }
  }
  for (int i = 0; i < n - 1; i++)
    res = max(res, max(dp1[i], oc[i]) + max(dp2[i + 1], tc[i + 1]));
  cout << res << endl;
  return 0;
}

### 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;
const int MAXN = 2005;
long long int n, a[MAXN], dp1[MAXN], dp2[MAXN], oc[MAXN], tc[MAXN], res;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  if (a[0] == 1)
    oc[0] = 1;
  else
    dp1[0] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == 1) {
      oc[i] = oc[i - 1] + 1;
      dp1[i] = dp1[i - 1];
    } else {
      dp1[i] = max(oc[i - 1], dp1[i - 1]) + 1;
      oc[i] = oc[i - 1];
    }
  }
  if (a[n - 1] == 2)
    tc[n - 1] = 1;
  else
    dp2[n - 1] = 1;
  for (int i = n - 2; i >= 0; i--) {
    if (a[i] == 1) {
      tc[i] = tc[i + 1];
      dp2[i] = max(tc[i + 1], dp2[i + 1]) + 1;
    } else {
      tc[i] = tc[i + 1] + 1;
      dp2[i] = dp2[i + 1];
    }
  }
  for (int i = 0; i < n - 1; i++)
    res = max(res, max(dp1[i], oc[i]) + max(dp2[i + 1], tc[i + 1]));
  cout << res << endl;
  return 0;
}

```