#include <bits/stdc++.h>
using namespace std;
const long long N = 2002;
const long long inf = 1e18;
long long a[N], ans[N][N][2], s[N], e[N];
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (a[0] == 1)
    s[0] = 1;
  else
    s[0] = 0;
  for (int i = 1; i < n; i++) {
    s[i] = s[i - 1];
    if (a[i] == 1) s[i]++;
  }
  if (a[n - 1] == 2)
    e[n - 1] = 1;
  else
    e[n - 1] = 0;
  for (int i = n - 2; i >= 0; i--) {
    e[i] = e[i + 1];
    if (a[i] == 2) e[i]++;
  }
  long long mx = max(s[n - 1], e[0]);
  for (int i = 0; i < n - 1; i++) mx = max(mx, s[i] + e[i + 1]);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1)
      a[i]++;
    else
      a[i]--;
  }
  for (int i = 0; i < n; i++)
    if (a[i] == 1) {
      ans[i][i][0] = 1;
      ans[i][i][1] = 1;
    } else {
      ans[i][i][0] = 0;
      ans[i][i][1] = 1;
    }
  for (int len = 2; len <= n; len++) {
    for (int i = 0; i < n; i++) {
      long long j = i + len - 1;
      if (j >= n) break;
      ans[i][j][0] = ans[i][j - 1][0];
      if (a[j] == 1) ans[i][j][0]++;
      ans[i][j][1] = ans[i][j - 1][1];
      if (a[j] == 2) ans[i][j][1]++;
      ans[i][j][1] = max(ans[i][j][0], ans[i][j][1]);
      long long val = ans[i][j][1];
      if (i != 0) val += s[i - 1];
      if (j != n) val += e[j + 1];
      mx = max(mx, val);
    }
  }
  cout << mx << endl;
}

### 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;
const long long N = 2002;
const long long inf = 1e18;
long long a[N], ans[N][N][2], s[N], e[N];
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (a[0] == 1)
    s[0] = 1;
  else
    s[0] = 0;
  for (int i = 1; i < n; i++) {
    s[i] = s[i - 1];
    if (a[i] == 1) s[i]++;
  }
  if (a[n - 1] == 2)
    e[n - 1] = 1;
  else
    e[n - 1] = 0;
  for (int i = n - 2; i >= 0; i--) {
    e[i] = e[i + 1];
    if (a[i] == 2) e[i]++;
  }
  long long mx = max(s[n - 1], e[0]);
  for (int i = 0; i < n - 1; i++) mx = max(mx, s[i] + e[i + 1]);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1)
      a[i]++;
    else
      a[i]--;
  }
  for (int i = 0; i < n; i++)
    if (a[i] == 1) {
      ans[i][i][0] = 1;
      ans[i][i][1] = 1;
    } else {
      ans[i][i][0] = 0;
      ans[i][i][1] = 1;
    }
  for (int len = 2; len <= n; len++) {
    for (int i = 0; i < n; i++) {
      long long j = i + len - 1;
      if (j >= n) break;
      ans[i][j][0] = ans[i][j - 1][0];
      if (a[j] == 1) ans[i][j][0]++;
      ans[i][j][1] = ans[i][j - 1][1];
      if (a[j] == 2) ans[i][j][1]++;
      ans[i][j][1] = max(ans[i][j][0], ans[i][j][1]);
      long long val = ans[i][j][1];
      if (i != 0) val += s[i - 1];
      if (j != n) val += e[j + 1];
      mx = max(mx, val);
    }
  }
  cout << mx << endl;
}

```