#include <bits/stdc++.h>
using namespace std;
int sum[4000];
int down[2001][2001][3];
int get_sum(int l, int r, int val) {
  if (l > r) return 0;
  int s = sum[r];
  if (l != 0) s -= sum[l - 1];
  if (val == 2)
    return (r - l + 1) - s;
  else
    return s;
}
int main() {
  int n, a[3000];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i != 0) sum[i] = sum[i - 1];
    sum[i] += (a[i] - 1) ^ 1;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 2)
      down[i][i][2] = 1;
    else
      down[i][i][1] = 1;
    for (int j = i + 1; j < n; j++) {
      if (a[j] == 1) {
        down[i][j][2] = down[i][j - 1][2];
        down[i][j][1] = max(down[i][j - 1][1], down[i][j - 1][2]) + 1;
      }
      if (a[j] == 2) {
        down[i][j][2] = down[i][j - 1][2] + 1;
        down[i][j][1] = down[i][j - 1][1];
      }
    }
  }
  int answer = 0;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      bool b1 = false, b2 = false;
      if (get_sum(i, j, 1) != 0) {
        b1 = true;
        int s = sum[j];
        answer = max(answer, s + get_sum(j + 1, n - 1, 1));
        answer = max(answer, s + get_sum(j + 1, n - 1, 2));
      }
      if ((i != 0 && (j - i + 1) - (sum[j] - sum[i - 1]) != 0) ||
          (i == 0 && (j + 1) - sum[j] != 0)) {
        b2 = true;
        int s = get_sum(i, n - 1, 2);
        answer = max(answer, get_sum(0, i - 1, 1) + s);
        answer = max(answer, get_sum(0, i - 1, 2) + s);
      }
      if (b1 && b2) {
        answer = max(answer, get_sum(0, i - 1, 1) + down[i][j][1] +
                                 get_sum(j + 1, n - 1, 2));
      }
    }
  cout << answer;
  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>
using namespace std;
int sum[4000];
int down[2001][2001][3];
int get_sum(int l, int r, int val) {
  if (l > r) return 0;
  int s = sum[r];
  if (l != 0) s -= sum[l - 1];
  if (val == 2)
    return (r - l + 1) - s;
  else
    return s;
}
int main() {
  int n, a[3000];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i != 0) sum[i] = sum[i - 1];
    sum[i] += (a[i] - 1) ^ 1;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 2)
      down[i][i][2] = 1;
    else
      down[i][i][1] = 1;
    for (int j = i + 1; j < n; j++) {
      if (a[j] == 1) {
        down[i][j][2] = down[i][j - 1][2];
        down[i][j][1] = max(down[i][j - 1][1], down[i][j - 1][2]) + 1;
      }
      if (a[j] == 2) {
        down[i][j][2] = down[i][j - 1][2] + 1;
        down[i][j][1] = down[i][j - 1][1];
      }
    }
  }
  int answer = 0;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      bool b1 = false, b2 = false;
      if (get_sum(i, j, 1) != 0) {
        b1 = true;
        int s = sum[j];
        answer = max(answer, s + get_sum(j + 1, n - 1, 1));
        answer = max(answer, s + get_sum(j + 1, n - 1, 2));
      }
      if ((i != 0 && (j - i + 1) - (sum[j] - sum[i - 1]) != 0) ||
          (i == 0 && (j + 1) - sum[j] != 0)) {
        b2 = true;
        int s = get_sum(i, n - 1, 2);
        answer = max(answer, get_sum(0, i - 1, 1) + s);
        answer = max(answer, get_sum(0, i - 1, 2) + s);
      }
      if (b1 && b2) {
        answer = max(answer, get_sum(0, i - 1, 1) + down[i][j][1] +
                                 get_sum(j + 1, n - 1, 2));
      }
    }
  cout << answer;
  return 0;
}

```