#include <bits/stdc++.h>
using namespace std;
int TM;
int n, v[2003], dp[2003][3][4][3][3];
int f(int in, int ls, int md, int ul, int dl) {
  if (in == n) return 0;
  if (dp[in][ls][md][ul][dl] != -1) return dp[in][ls][md][ul][dl];
  int res = 0;
  if (md == 1) {
    if (v[in] >= ls) {
      res = max(res, 1 + f(in + 1, v[in], 1, 0, v[in]));
      res = max(res, 1 + f(in + 1, v[in], 2, v[in], dl));
    }
    res = max(res, 0 + f(in + 1, ls, 1, 0, dl));
  }
  if (md == 2) {
    if (v[in] <= ls && v[in] >= dl)
      res = max(res, 1 + f(in + 1, v[in], 2, ul, dl));
    if (v[in] >= ul) res = max(res, 1 + f(in + 1, v[in], 3, ul, dl));
    res = max(res, 0 + f(in + 1, ls, 2, ul, dl));
  }
  if (md == 3) {
    if (v[in] >= ls) res = max(res, 1 + f(in + 1, v[in], 3, ul, dl));
    res = max(res, 0 + f(in + 1, ls, 3, ul, dl));
  }
  return dp[in][ls][md][ul][dl] = res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  int ans = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) ans = max(ans, f(i, v[i], 1, 0, 0));
  cout << ans;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 TM;
int n, v[2003], dp[2003][3][4][3][3];
int f(int in, int ls, int md, int ul, int dl) {
  if (in == n) return 0;
  if (dp[in][ls][md][ul][dl] != -1) return dp[in][ls][md][ul][dl];
  int res = 0;
  if (md == 1) {
    if (v[in] >= ls) {
      res = max(res, 1 + f(in + 1, v[in], 1, 0, v[in]));
      res = max(res, 1 + f(in + 1, v[in], 2, v[in], dl));
    }
    res = max(res, 0 + f(in + 1, ls, 1, 0, dl));
  }
  if (md == 2) {
    if (v[in] <= ls && v[in] >= dl)
      res = max(res, 1 + f(in + 1, v[in], 2, ul, dl));
    if (v[in] >= ul) res = max(res, 1 + f(in + 1, v[in], 3, ul, dl));
    res = max(res, 0 + f(in + 1, ls, 2, ul, dl));
  }
  if (md == 3) {
    if (v[in] >= ls) res = max(res, 1 + f(in + 1, v[in], 3, ul, dl));
    res = max(res, 0 + f(in + 1, ls, 3, ul, dl));
  }
  return dp[in][ls][md][ul][dl] = res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  int ans = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) ans = max(ans, f(i, v[i], 1, 0, 0));
  cout << ans;
}

```