#include <bits/stdc++.h>
#pragma optimize GCC("Ofast")
#pragma comment(linker, "/STACK:1073741824")
using namespace std;
using namespace placeholders;
const int inf = (int)1e9 + 9;
const long long linf = (long long)1e18;
const long double eps = 1e-7;
const int mod = (int)1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  srand(time(0));
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> seq(n);
  for (int i = 0; i < n; i++) {
    cin >> seq[i];
  }
  vector<int> cnt1(n + 1), cnt2(n + 1);
  for (int i = 1; i <= n; i++) {
    cnt1[i] = cnt1[i - 1];
    if (seq[i - 1] == 1) {
      cnt1[i]++;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    cnt2[i] = cnt2[i + 1];
    if (seq[i] == 2) {
      cnt2[i]++;
    }
  }
  vector<vector<int> > dp(n + 2, vector<int>(n + 1, 0));
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 1;
    for (int j = i + 1; j <= n; j++) {
      if (seq[j - 1] == 1) {
        dp[i][j] = max(dp[i][j - 1], cnt1[j] - cnt1[i - 1]);
      } else {
        dp[i][j] = dp[i][j - 1] + 1;
      }
    }
  }
  vector<vector<int> > dp_rev(n + 2, vector<int>(n + 1, 0));
  vector<int> seq_rev(n);
  for (int i = 0; i < n; i++) {
    seq_rev[i] = 3 - seq[i];
  }
  for (int i = 1; i <= n; i++) {
    dp_rev[i][i] = 1;
    for (int j = i + 1; j <= n; j++) {
      if (seq_rev[j - 1] == 1) {
        dp_rev[i][j] =
            max(dp_rev[i][j - 1], j - i + 1 - (cnt1[j] - cnt1[i - 1]));
      } else {
        dp_rev[i][j] = dp_rev[i][j - 1] + 1;
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= n; i++) {
    int one = 0, two = 0;
    for (int j = i; j <= n; j++) {
      if (seq[j - 1] == 1) {
        one++;
      } else {
        two++;
      }
      res = max(res, dp[1][i - 1] + cnt2[j] + one);
      res = max(res, dp_rev[i][j] + cnt1[i - 1] + cnt2[j]);
      res = max(res, cnt1[i - 1] + two + dp[j + 1][n]);
    }
  }
  cout << res;
}

### Prompt

Generate a cpp solution to 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>
#pragma optimize GCC("Ofast")
#pragma comment(linker, "/STACK:1073741824")
using namespace std;
using namespace placeholders;
const int inf = (int)1e9 + 9;
const long long linf = (long long)1e18;
const long double eps = 1e-7;
const int mod = (int)1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  srand(time(0));
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> seq(n);
  for (int i = 0; i < n; i++) {
    cin >> seq[i];
  }
  vector<int> cnt1(n + 1), cnt2(n + 1);
  for (int i = 1; i <= n; i++) {
    cnt1[i] = cnt1[i - 1];
    if (seq[i - 1] == 1) {
      cnt1[i]++;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    cnt2[i] = cnt2[i + 1];
    if (seq[i] == 2) {
      cnt2[i]++;
    }
  }
  vector<vector<int> > dp(n + 2, vector<int>(n + 1, 0));
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 1;
    for (int j = i + 1; j <= n; j++) {
      if (seq[j - 1] == 1) {
        dp[i][j] = max(dp[i][j - 1], cnt1[j] - cnt1[i - 1]);
      } else {
        dp[i][j] = dp[i][j - 1] + 1;
      }
    }
  }
  vector<vector<int> > dp_rev(n + 2, vector<int>(n + 1, 0));
  vector<int> seq_rev(n);
  for (int i = 0; i < n; i++) {
    seq_rev[i] = 3 - seq[i];
  }
  for (int i = 1; i <= n; i++) {
    dp_rev[i][i] = 1;
    for (int j = i + 1; j <= n; j++) {
      if (seq_rev[j - 1] == 1) {
        dp_rev[i][j] =
            max(dp_rev[i][j - 1], j - i + 1 - (cnt1[j] - cnt1[i - 1]));
      } else {
        dp_rev[i][j] = dp_rev[i][j - 1] + 1;
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= n; i++) {
    int one = 0, two = 0;
    for (int j = i; j <= n; j++) {
      if (seq[j - 1] == 1) {
        one++;
      } else {
        two++;
      }
      res = max(res, dp[1][i - 1] + cnt2[j] + one);
      res = max(res, dp_rev[i][j] + cnt1[i - 1] + cnt2[j]);
      res = max(res, cnt1[i - 1] + two + dp[j + 1][n]);
    }
  }
  cout << res;
}

```