#include <bits/stdc++.h>
using namespace std;
int setbit(int n, int i, int v) {
  return ((v) ? (n | (1 << i)) : (n & ~(1 << i)));
}
long long gcd(int i, int j) {
  if (j == 0) return i;
  return gcd(j, i % j);
}
bool chk(int n, int i) {
  if ((n >> i) & 1 == 1) return 1;
  return 0;
}
long long max(long long a, long long b) { return a >= b ? a : b; }
string tostring1(long long n) {
  string res = "";
  while (n) {
    res += ((n % 10) + '0');
    n /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
bool T(pair<int, int> p1, pair<int, int> p2) {
  return (p2.second - p1.first >= p2.second - p2.first);
}
long long todec(string s) {
  long long res = 0;
  long long cnt = 0;
  for (int i = s.length() - 1; i >= 0; i--) {
    res += (s[i] - '0') * pow(10, cnt);
    cnt++;
  }
  return res;
}
int n;
int a[2003];
int mem[2003][5];
int dp(int i, int j) {
  if (j >= 4) return -1e9;
  if (i >= n) return 0;
  int& ret = mem[i][j];
  if (ret != -1) return ret;
  ret = 0;
  if (j == 0) {
    if (a[i] == 1) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else if (j == 1) {
    if (a[i] == 2) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else if (j == 2) {
    if (a[i] == 1) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else {
    if (a[i] == 2) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(mem, -1, sizeof mem);
  cout << max(dp(0, 0), dp(0, 1)) << endl;
  return 0;
}

### 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;
int setbit(int n, int i, int v) {
  return ((v) ? (n | (1 << i)) : (n & ~(1 << i)));
}
long long gcd(int i, int j) {
  if (j == 0) return i;
  return gcd(j, i % j);
}
bool chk(int n, int i) {
  if ((n >> i) & 1 == 1) return 1;
  return 0;
}
long long max(long long a, long long b) { return a >= b ? a : b; }
string tostring1(long long n) {
  string res = "";
  while (n) {
    res += ((n % 10) + '0');
    n /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
bool T(pair<int, int> p1, pair<int, int> p2) {
  return (p2.second - p1.first >= p2.second - p2.first);
}
long long todec(string s) {
  long long res = 0;
  long long cnt = 0;
  for (int i = s.length() - 1; i >= 0; i--) {
    res += (s[i] - '0') * pow(10, cnt);
    cnt++;
  }
  return res;
}
int n;
int a[2003];
int mem[2003][5];
int dp(int i, int j) {
  if (j >= 4) return -1e9;
  if (i >= n) return 0;
  int& ret = mem[i][j];
  if (ret != -1) return ret;
  ret = 0;
  if (j == 0) {
    if (a[i] == 1) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else if (j == 1) {
    if (a[i] == 2) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else if (j == 2) {
    if (a[i] == 1) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  } else {
    if (a[i] == 2) {
      ret = dp(i + 1, j + 1) + 1;
      ret = max(ret, dp(i + 1, j) + 1);
      ret = max(ret, dp(i + 1, j));
    } else
      ret = dp(i + 1, j);
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(mem, -1, sizeof mem);
  cout << max(dp(0, 0), dp(0, 1)) << endl;
  return 0;
}

```