#include <bits/stdc++.h>
namespace in {
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int getc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <typename T>
inline void read(T& t) {
  t = 0;
  int f = 0;
  char ch = getc();
  while (!isdigit(ch)) {
    if (ch == '-') f = 1;
    ch = getc();
  }
  while (isdigit(ch)) {
    t = t * 10 + ch - 48;
    ch = getc();
  }
  if (f) t = -t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
}  // namespace in
namespace out {
char buffer[1 << 21];
int p1 = -1;
const int p2 = (1 << 21) - 1;
inline void flush() { fwrite(buffer, 1, p1 + 1, stdout), p1 = -1; }
inline void putc(const char& x) {
  if (p1 == p2) flush();
  buffer[++p1] = x;
}
template <typename T>
void write(T x) {
  static char buf[15];
  static int len = -1;
  if (x >= 0) {
    do {
      buf[++len] = x % 10 + 48, x /= 10;
    } while (x);
  } else {
    putc('-');
    do {
      buf[++len] = -(x % 10) + 48, x /= 10;
    } while (x);
  }
  while (len >= 0) putc(buf[len]), --len;
}
}  // namespace out
using namespace std;
const int N = 2e3 + 10;
int n, a[N], f[N][5];
signed main() {
  in::read(n);
  for (int i = 1; i <= n; i++) in::read(a[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 4; j++) {
      f[i][j] = f[i][j - 1];
      if (a[i] == (j + 1) % 2 + 1)
        (f[i][j]) = max((f[i][j]), (f[i - 1][j] + 1));
      else
        (f[i][j]) = max((f[i][j]), (f[i - 1][j]));
    }
  }
  out::write(f[n][4]);
  out::flush();
  return 0;
}

### Prompt

Your challenge is to write 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>
namespace in {
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int getc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <typename T>
inline void read(T& t) {
  t = 0;
  int f = 0;
  char ch = getc();
  while (!isdigit(ch)) {
    if (ch == '-') f = 1;
    ch = getc();
  }
  while (isdigit(ch)) {
    t = t * 10 + ch - 48;
    ch = getc();
  }
  if (f) t = -t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
}  // namespace in
namespace out {
char buffer[1 << 21];
int p1 = -1;
const int p2 = (1 << 21) - 1;
inline void flush() { fwrite(buffer, 1, p1 + 1, stdout), p1 = -1; }
inline void putc(const char& x) {
  if (p1 == p2) flush();
  buffer[++p1] = x;
}
template <typename T>
void write(T x) {
  static char buf[15];
  static int len = -1;
  if (x >= 0) {
    do {
      buf[++len] = x % 10 + 48, x /= 10;
    } while (x);
  } else {
    putc('-');
    do {
      buf[++len] = -(x % 10) + 48, x /= 10;
    } while (x);
  }
  while (len >= 0) putc(buf[len]), --len;
}
}  // namespace out
using namespace std;
const int N = 2e3 + 10;
int n, a[N], f[N][5];
signed main() {
  in::read(n);
  for (int i = 1; i <= n; i++) in::read(a[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 4; j++) {
      f[i][j] = f[i][j - 1];
      if (a[i] == (j + 1) % 2 + 1)
        (f[i][j]) = max((f[i][j]), (f[i - 1][j] + 1));
      else
        (f[i][j]) = max((f[i][j]), (f[i - 1][j]));
    }
  }
  out::write(f[n][4]);
  out::flush();
  return 0;
}

```