#include <bits/stdc++.h>
using namespace std;
const int OO = 0x3f3f3f3f;
const double eps = (1e-10);
int n, a[20005], mem[2000][4], ans;
int dp(int idx, int state) {
  if (idx == n) return 0;
  int& ret = mem[idx][state];
  if (ret != -1) return ret;
  ret = dp(idx + 1, state);
  if ((a[idx] == 1 && (state == 0 || state == 2)) ||
      (a[idx] == 2 && (state == 1 || state == 3)))
    ret = max(ret, 1 + dp(idx + 1, state));
  if ((a[idx] == 1 && state == 1) ||
      (a[idx] == 2 && (state == 0 || state == 2)))
    ret = max(ret, 1 + dp(idx + 1, state + 1));
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(7);
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  memset(mem, -1, sizeof(mem));
  cout << dp(0, 0);
  return 0;
}

### Prompt

Develop a solution in Cpp 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 int OO = 0x3f3f3f3f;
const double eps = (1e-10);
int n, a[20005], mem[2000][4], ans;
int dp(int idx, int state) {
  if (idx == n) return 0;
  int& ret = mem[idx][state];
  if (ret != -1) return ret;
  ret = dp(idx + 1, state);
  if ((a[idx] == 1 && (state == 0 || state == 2)) ||
      (a[idx] == 2 && (state == 1 || state == 3)))
    ret = max(ret, 1 + dp(idx + 1, state));
  if ((a[idx] == 1 && state == 1) ||
      (a[idx] == 2 && (state == 0 || state == 2)))
    ret = max(ret, 1 + dp(idx + 1, state + 1));
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(7);
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  memset(mem, -1, sizeof(mem));
  cout << dp(0, 0);
  return 0;
}

```