#include <bits/stdc++.h>
template <typename tp>
inline tp max(tp x, tp y) {
  return (x > y) ? x : y;
}
template <typename tp>
inline tp min(tp x, tp y) {
  return (x < y) ? x : y;
}
template <typename tp>
inline bool chkmax(tp& x, tp y) {
  return (x >= y) ? false : (x = y, true);
}
template <typename tp>
inline bool chkmin(tp& x, tp y) {
  return (x <= y) ? false : (x = y, true);
}
template <typename tp>
tp read(tp& in) {
  in = 0;
  char ch = getchar();
  tp f = 1;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) in = in * 10 + ch - '0';
  return in *= f;
}
static const int max1 = 2000 + 11;
int n;
int ai[max1], S[max1][4];
int f[max1];
int main() {
  read(n);
  for (int i = (1), ir = (n); i <= (ir); ++i) read(ai[i]);
  for (int i = (1), ir = (n); i <= (ir); ++i) {
    S[i][0] = S[i - 1][0], S[i][1] = S[i - 1][1];
    if (ai[i] == 1)
      ++S[i][0];
    else
      ++S[i][1];
  }
  for (int i = (n), ir = (1); i >= (ir); --i) {
    S[i][2] = S[i + 1][2], S[i][3] = S[i + 1][3];
    if (ai[i] == 1)
      ++S[i][2];
    else
      ++S[i][3];
  }
  for (int i = (n), ir = (1); i >= (ir); --i) {
    int temp = 0;
    for (int j = (i), jr = (n); j <= (jr); ++j) {
      if (ai[j] == 1) ++temp;
      chkmax(f[i], temp + S[j][3]);
    }
  }
  int ans = 0;
  for (int i = (1), ir = (n); i <= (ir); ++i) {
    int temp = 0;
    for (int j = (i), jr = (n); j <= (jr); ++j) {
      if (ai[j] == 2) ++temp;
      chkmax(ans, S[i][0] + temp + f[j + 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
template <typename tp>
inline tp max(tp x, tp y) {
  return (x > y) ? x : y;
}
template <typename tp>
inline tp min(tp x, tp y) {
  return (x < y) ? x : y;
}
template <typename tp>
inline bool chkmax(tp& x, tp y) {
  return (x >= y) ? false : (x = y, true);
}
template <typename tp>
inline bool chkmin(tp& x, tp y) {
  return (x <= y) ? false : (x = y, true);
}
template <typename tp>
tp read(tp& in) {
  in = 0;
  char ch = getchar();
  tp f = 1;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) in = in * 10 + ch - '0';
  return in *= f;
}
static const int max1 = 2000 + 11;
int n;
int ai[max1], S[max1][4];
int f[max1];
int main() {
  read(n);
  for (int i = (1), ir = (n); i <= (ir); ++i) read(ai[i]);
  for (int i = (1), ir = (n); i <= (ir); ++i) {
    S[i][0] = S[i - 1][0], S[i][1] = S[i - 1][1];
    if (ai[i] == 1)
      ++S[i][0];
    else
      ++S[i][1];
  }
  for (int i = (n), ir = (1); i >= (ir); --i) {
    S[i][2] = S[i + 1][2], S[i][3] = S[i + 1][3];
    if (ai[i] == 1)
      ++S[i][2];
    else
      ++S[i][3];
  }
  for (int i = (n), ir = (1); i >= (ir); --i) {
    int temp = 0;
    for (int j = (i), jr = (n); j <= (jr); ++j) {
      if (ai[j] == 1) ++temp;
      chkmax(f[i], temp + S[j][3]);
    }
  }
  int ans = 0;
  for (int i = (1), ir = (n); i <= (ir); ++i) {
    int temp = 0;
    for (int j = (i), jr = (n); j <= (jr); ++j) {
      if (ai[j] == 2) ++temp;
      chkmax(ans, S[i][0] + temp + f[j + 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```