#include <bits/stdc++.h>
using namespace std;
int n;
int a[10001], sum1[10001], sum2[10001], f[2001][2001], g[2001][2001];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    sum1[i] = sum1[i - 1] + (a[i] == 1);
    sum2[i] = sum2[i - 1] + (a[i] == 2);
  }
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= i; j++)
      if (a[j] == 2) now++;
    f[i][i] = now;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 2) now++;
      if (a[j - i] == 2) now--;
      f[j][i] = max(f[j - 1][i], now);
    }
  }
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = n; j >= n - i + 1; j--)
      if (a[j] == 1) now++;
    g[n - i + 1][i] = now;
    for (int j = n - i; j >= 1; j--) {
      if (a[j] == 1) now++;
      if (a[j + i] == 1) now--;
      g[j][i] = max(g[j + 1][i], now);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    ans = max(ans, sum1[i] + sum2[n] - sum2[i]);
    for (int j = 1; j <= i; j++) {
      if (n - i < i - j + 1) continue;
      int k1 = g[i + 1][i - j + 1];
      int k2 = sum2[i] - sum2[j - 1];
      ans = max(ans, sum1[i] + sum2[n] - sum2[i] - ((i - j + 1) * 2 - k1 - k2) +
                         k1 + k2);
    }
    for (int j = i + 1; j <= n; j++) {
      if (j - i > i) continue;
      int k1 = sum1[j] - sum1[i];
      int k2 = f[i][j - i];
      ans = max(
          ans, sum1[i] + sum2[n] - sum2[i] - ((j - i) * 2 - k1 - k2) + k1 + k2);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 n;
int a[10001], sum1[10001], sum2[10001], f[2001][2001], g[2001][2001];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    sum1[i] = sum1[i - 1] + (a[i] == 1);
    sum2[i] = sum2[i - 1] + (a[i] == 2);
  }
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= i; j++)
      if (a[j] == 2) now++;
    f[i][i] = now;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 2) now++;
      if (a[j - i] == 2) now--;
      f[j][i] = max(f[j - 1][i], now);
    }
  }
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = n; j >= n - i + 1; j--)
      if (a[j] == 1) now++;
    g[n - i + 1][i] = now;
    for (int j = n - i; j >= 1; j--) {
      if (a[j] == 1) now++;
      if (a[j + i] == 1) now--;
      g[j][i] = max(g[j + 1][i], now);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    ans = max(ans, sum1[i] + sum2[n] - sum2[i]);
    for (int j = 1; j <= i; j++) {
      if (n - i < i - j + 1) continue;
      int k1 = g[i + 1][i - j + 1];
      int k2 = sum2[i] - sum2[j - 1];
      ans = max(ans, sum1[i] + sum2[n] - sum2[i] - ((i - j + 1) * 2 - k1 - k2) +
                         k1 + k2);
    }
    for (int j = i + 1; j <= n; j++) {
      if (j - i > i) continue;
      int k1 = sum1[j] - sum1[i];
      int k2 = f[i][j - i];
      ans = max(
          ans, sum1[i] + sum2[n] - sum2[i] - ((j - i) * 2 - k1 - k2) + k1 + k2);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```