#include <bits/stdc++.h>
using namespace std;
const int N = 2003;
int n, res, a[N], f[N][N], g[3][N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int len = 1; len <= n; ++len) {
    int one = 0;
    int two = 0;
    for (int i = 1; i + len - 1 <= n; ++i) {
      int j = i + len - 1;
      if (i == 1) {
        for (int k = i; k < j; ++k) {
          if (a[k] == 1)
            one = k;
          else
            two = k;
        }
      }
      if (a[j] == 2)
        f[i][j] = f[i][two] + 1, two = j;
      else
        f[i][j] = max(f[i][one], f[i][two]) + 1, one = j;
    }
  }
  for (int i = 1; i <= n; ++i) {
    g[1][i] = g[1][i - 1];
    g[2][i] = g[2][i - 1];
    if (a[i] == 1)
      ++g[1][i];
    else
      ++g[2][i];
  }
  int i0, j0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int k = g[1][i - 1] + f[i][j] + g[2][n] - g[2][j];
      if (k > res) res = k;
    }
  }
  res = max(res, g[1][n]);
  res = max(res, g[2][n]);
  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 = 2003;
int n, res, a[N], f[N][N], g[3][N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int len = 1; len <= n; ++len) {
    int one = 0;
    int two = 0;
    for (int i = 1; i + len - 1 <= n; ++i) {
      int j = i + len - 1;
      if (i == 1) {
        for (int k = i; k < j; ++k) {
          if (a[k] == 1)
            one = k;
          else
            two = k;
        }
      }
      if (a[j] == 2)
        f[i][j] = f[i][two] + 1, two = j;
      else
        f[i][j] = max(f[i][one], f[i][two]) + 1, one = j;
    }
  }
  for (int i = 1; i <= n; ++i) {
    g[1][i] = g[1][i - 1];
    g[2][i] = g[2][i - 1];
    if (a[i] == 1)
      ++g[1][i];
    else
      ++g[2][i];
  }
  int i0, j0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int k = g[1][i - 1] + f[i][j] + g[2][n] - g[2][j];
      if (k > res) res = k;
    }
  }
  res = max(res, g[1][n]);
  res = max(res, g[2][n]);
  printf("%d\n", res);
  return 0;
}

```