#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const long long M = (long long)1e9 + 7;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void Char(char *ch) { scanf("%s", ch); }
template <typename t>
inline t abs(t a) {
  if (a >= 0) return a;
  return -a;
}
template <typename T>
inline T modpow(T b, T p, T Mod) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
      res %= Mod;
    }
    b *= b;
    b %= Mod;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T ModInv(T b, T Mod) {
  return modpow(b, Mod - 2, Mod);
}
template <typename T>
inline T pwr(T b, T p) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
    }
    b *= b;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename U>
inline void Int(T &a, U &b) {
  Int(a);
  Int(b);
}
template <typename T, typename U, typename W>
inline void Int(T &a, U &b, W &c) {
  Int(a, b);
  Int(c);
}
template <typename T, typename U, typename W, typename X>
inline void Int(T &a, U &b, W &c, X &d) {
  Int(a, b, c);
  Int(d);
}
vector<int> Unique(vector<int> v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  return v;
}
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cout << it->substr((*it)[0] == ' ', it->length()) << " = " << a << " ";
  err(++it, args...);
}
int a[N], dp[3][2005], dpr[2005][2005];
int main() {
  int n = Int();
  for (int i = 1; i <= n; i++) {
    a[i] = Int();
    if (a[i] == 1) {
      dp[1][i] = 1 + dp[1][i - 1];
      dp[2][i] = dp[2][i - 1];
    } else {
      dp[2][i] = 1 + dp[2][i - 1];
      dp[1][i] = dp[1][i - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    int one = 0, two = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 2)
        two++;
      else {
        one = max(one, two) + 1;
      }
      dpr[i][j] = max(one, two);
    }
  }
  int res = max(dp[1][n], dp[2][n]);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      res = max(res, dp[1][i - 1] + dp[2][n] - dp[2][j] + dpr[i][j]);
    }
  }
  printf("%d\n", res);
  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;
const int N = 2e5 + 5;
const long long M = (long long)1e9 + 7;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void Char(char *ch) { scanf("%s", ch); }
template <typename t>
inline t abs(t a) {
  if (a >= 0) return a;
  return -a;
}
template <typename T>
inline T modpow(T b, T p, T Mod) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
      res %= Mod;
    }
    b *= b;
    b %= Mod;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T ModInv(T b, T Mod) {
  return modpow(b, Mod - 2, Mod);
}
template <typename T>
inline T pwr(T b, T p) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
    }
    b *= b;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename U>
inline void Int(T &a, U &b) {
  Int(a);
  Int(b);
}
template <typename T, typename U, typename W>
inline void Int(T &a, U &b, W &c) {
  Int(a, b);
  Int(c);
}
template <typename T, typename U, typename W, typename X>
inline void Int(T &a, U &b, W &c, X &d) {
  Int(a, b, c);
  Int(d);
}
vector<int> Unique(vector<int> v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  return v;
}
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cout << it->substr((*it)[0] == ' ', it->length()) << " = " << a << " ";
  err(++it, args...);
}
int a[N], dp[3][2005], dpr[2005][2005];
int main() {
  int n = Int();
  for (int i = 1; i <= n; i++) {
    a[i] = Int();
    if (a[i] == 1) {
      dp[1][i] = 1 + dp[1][i - 1];
      dp[2][i] = dp[2][i - 1];
    } else {
      dp[2][i] = 1 + dp[2][i - 1];
      dp[1][i] = dp[1][i - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    int one = 0, two = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 2)
        two++;
      else {
        one = max(one, two) + 1;
      }
      dpr[i][j] = max(one, two);
    }
  }
  int res = max(dp[1][n], dp[2][n]);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      res = max(res, dp[1][i - 1] + dp[2][n] - dp[2][j] + dpr[i][j]);
    }
  }
  printf("%d\n", res);
  return 0;
}

```