#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const int MAX = 100005;
int a[2002];
int fdp[2002][2002][2];
int rdp[2002][2002][2];
int fsolve(int b, int e, int last) {
  if (b > e) return 0;
  int &ret = fdp[b][e][last];
  if (ret != -1) return ret;
  ret = fsolve(b + 1, e, last);
  if (last <= a[b]) ret = max(ret, 1 + fsolve(b + 1, e, a[b]));
  return ret;
}
int rsolve(int b, int e, int last) {
  if (b > e) return 0;
  int &ret = rdp[b][e][last];
  if (ret != -1) return ret;
  ret = rsolve(b, e - 1, last);
  if (last <= a[e]) ret = max(ret, 1 + rsolve(b, e - 1, a[e]));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], a[i]--;
  memset(fdp, -1, sizeof(fdp));
  memset(rdp, -1, sizeof(rdp));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 0) + fsolve(j + 1, n, 0));
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 0) + fsolve(j + 1, n, 1));
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 1) + fsolve(j + 1, n, 1));
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
using namespace std;
const long long MOD = 1000000007;
const int MAX = 100005;
int a[2002];
int fdp[2002][2002][2];
int rdp[2002][2002][2];
int fsolve(int b, int e, int last) {
  if (b > e) return 0;
  int &ret = fdp[b][e][last];
  if (ret != -1) return ret;
  ret = fsolve(b + 1, e, last);
  if (last <= a[b]) ret = max(ret, 1 + fsolve(b + 1, e, a[b]));
  return ret;
}
int rsolve(int b, int e, int last) {
  if (b > e) return 0;
  int &ret = rdp[b][e][last];
  if (ret != -1) return ret;
  ret = rsolve(b, e - 1, last);
  if (last <= a[e]) ret = max(ret, 1 + rsolve(b, e - 1, a[e]));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], a[i]--;
  memset(fdp, -1, sizeof(fdp));
  memset(rdp, -1, sizeof(rdp));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 0) + fsolve(j + 1, n, 0));
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 0) + fsolve(j + 1, n, 1));
      ans =
          max(ans, fsolve(1, i - 1, 0) + rsolve(i, j, 1) + fsolve(j + 1, n, 1));
    }
  }
  cout << ans << endl;
  return 0;
}

```