#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, a[N], pref[N], suf[N], dp[N][N][2][2];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] &= 1;
  }
  for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] + a[i];
  for (int i = n; i > 0; --i) suf[i] = suf[i + 1] + (a[i] ^ 1);
  for (int r = 1; r <= n; ++r) {
    dp[r][r][a[r]][a[r]] = 1;
    for (int l = r - 1; l > 0; --l) {
      dp[l][r][0][0] = max(dp[l][r][0][0], dp[l + 1][r][0][0] + (a[l] == 0));
      dp[l][r][1][1] = max(dp[l][r][1][1], dp[l + 1][r][1][1] + (a[l] == 1));
      dp[l][r][0][1] = max(dp[l][r][0][1], dp[l + 1][r][1][1] + (a[l] == 0));
      dp[l][r][0][1] = max(dp[l][r][0][1], dp[l + 1][r][0][1] + (a[l] == 0));
    }
  }
  int ans = 1;
  for (int l = 1; l <= n; ++l)
    for (int r = l; r <= n; ++r) {
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][0][1]);
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][1][1]);
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][0][0]);
    }
  cout << ans << endl;
}

### 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 int N = 2005;
int n, a[N], pref[N], suf[N], dp[N][N][2][2];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] &= 1;
  }
  for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] + a[i];
  for (int i = n; i > 0; --i) suf[i] = suf[i + 1] + (a[i] ^ 1);
  for (int r = 1; r <= n; ++r) {
    dp[r][r][a[r]][a[r]] = 1;
    for (int l = r - 1; l > 0; --l) {
      dp[l][r][0][0] = max(dp[l][r][0][0], dp[l + 1][r][0][0] + (a[l] == 0));
      dp[l][r][1][1] = max(dp[l][r][1][1], dp[l + 1][r][1][1] + (a[l] == 1));
      dp[l][r][0][1] = max(dp[l][r][0][1], dp[l + 1][r][1][1] + (a[l] == 0));
      dp[l][r][0][1] = max(dp[l][r][0][1], dp[l + 1][r][0][1] + (a[l] == 0));
    }
  }
  int ans = 1;
  for (int l = 1; l <= n; ++l)
    for (int r = l; r <= n; ++r) {
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][0][1]);
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][1][1]);
      ans = max(ans, pref[l - 1] + suf[r + 1] + dp[l][r][0][0]);
    }
  cout << ans << endl;
}

```