#include <bits/stdc++.h>
using namespace std;
int main() {
  int N;
  cin >> N;
  int v[N + 1], dp[N + 1], pref[N + 1];
  memset(dp, 0, sizeof(dp));
  memset(pref, 0, sizeof(pref));
  for (int i = 1; i <= N; ++i) {
    cin >> v[i];
    pref[i] = pref[i - 1];
    if (v[i] == 1) {
      ++pref[i];
    }
  }
  int ans = 1;
  for (int i = 1; i <= N; ++i) {
    dp[i] = 1;
    for (int j = i - 1; j >= 1; --j) {
      if (v[i] >= v[j] && 1 + dp[j] > dp[i]) {
        dp[i] = 1 + dp[j];
      }
    }
    if (dp[i] > ans) {
      ans = dp[i];
    }
  }
  int pref2 = 0;
  for (int i = N; i >= 1; --i) {
    if (v[i] == 2) {
      ++pref2;
    } else {
      int n1 = 0, tans = 0;
      for (int j = i; j >= 1; --j) {
        if (v[j] == 1) {
          ++n1;
        } else {
          tans = max(tans + 1, 1 + n1);
          if (tans + pref2 + pref[j - 1] > ans) {
            ans = tans + pref2 + pref[j - 1];
          }
        }
      }
    }
  }
  cout << ans << "\n";
}

### 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;
int main() {
  int N;
  cin >> N;
  int v[N + 1], dp[N + 1], pref[N + 1];
  memset(dp, 0, sizeof(dp));
  memset(pref, 0, sizeof(pref));
  for (int i = 1; i <= N; ++i) {
    cin >> v[i];
    pref[i] = pref[i - 1];
    if (v[i] == 1) {
      ++pref[i];
    }
  }
  int ans = 1;
  for (int i = 1; i <= N; ++i) {
    dp[i] = 1;
    for (int j = i - 1; j >= 1; --j) {
      if (v[i] >= v[j] && 1 + dp[j] > dp[i]) {
        dp[i] = 1 + dp[j];
      }
    }
    if (dp[i] > ans) {
      ans = dp[i];
    }
  }
  int pref2 = 0;
  for (int i = N; i >= 1; --i) {
    if (v[i] == 2) {
      ++pref2;
    } else {
      int n1 = 0, tans = 0;
      for (int j = i; j >= 1; --j) {
        if (v[j] == 1) {
          ++n1;
        } else {
          tans = max(tans + 1, 1 + n1);
          if (tans + pref2 + pref[j - 1] > ans) {
            ans = tans + pref2 + pref[j - 1];
          }
        }
      }
    }
  }
  cout << ans << "\n";
}

```