#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int N = 2e3 + 5;
int pre[N], en[N], a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  en[n + 1] = pre[0] = 0;
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; i--) en[i] = en[i + 1] + (a[i] == 2);
  int cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    cnt1 = cnt2 = 0;
    for (int j = i; j <= n; j++) {
      cnt2 += (a[j] == 2);
      cnt1 = max(cnt2, cnt1) + (a[j] == 1);
      ans = max(ans, max(cnt1, cnt2) + pre[i - 1] + en[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int N = 2e3 + 5;
int pre[N], en[N], a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  en[n + 1] = pre[0] = 0;
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; i--) en[i] = en[i + 1] + (a[i] == 2);
  int cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    cnt1 = cnt2 = 0;
    for (int j = i; j <= n; j++) {
      cnt2 += (a[j] == 2);
      cnt1 = max(cnt2, cnt1) + (a[j] == 1);
      ans = max(ans, max(cnt1, cnt2) + pre[i - 1] + en[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}

```