#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 5;
const int inf = 1e9 + 1;
const int mod = 1e9 + 7;
const double eps = 1e-15;
const int pw = 257;
int n, a[N], pr[N], sf[N], ans, cnt, f, s, b[N][N], c[N][N];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) pr[i] = pr[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; --i) sf[i] = sf[i + 1] + (a[i] == 2);
  for (int i = 0; i <= n; ++i) ans = max(ans, pr[i] + sf[i + 1]);
  for (int i = 0; i <= n + 1; ++i)
    for (int j = 1; j <= n; ++j) b[i][j] = c[i][j] = -inf;
  for (int i = n; i >= 1; --i) {
    cnt = f = s = 0;
    for (int j = 1; j <= n; ++j) {
      if (i - j + 1 == 0) break;
      if (a[i - j + 1] == 2) ++cnt;
      if (a[i - j + 1] == 2)
        ++s;
      else
        ++f;
      b[i - j + 1][j] = max(b[i - j + 2][j], f - s);
      if (i + j <= n) ans = max(ans, pr[i - j] + cnt + sf[i + 1] + b[i + 1][j]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    cnt = f = s = 0;
    for (int j = 1; j <= n; ++j) {
      if (i + j - 1 == n + 1) break;
      if (a[i + j - 1] == 1) ++cnt;
      if (a[i + j - 1] == 1)
        ++f;
      else
        ++s;
      c[i][j] = max(c[i - 1][j], s - f);
      if (i - j >= 1) ans = max(ans, sf[i + j] + pr[i - 1] + cnt + c[i - j][j]);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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 int N = 2e3 + 5;
const int inf = 1e9 + 1;
const int mod = 1e9 + 7;
const double eps = 1e-15;
const int pw = 257;
int n, a[N], pr[N], sf[N], ans, cnt, f, s, b[N][N], c[N][N];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) pr[i] = pr[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; --i) sf[i] = sf[i + 1] + (a[i] == 2);
  for (int i = 0; i <= n; ++i) ans = max(ans, pr[i] + sf[i + 1]);
  for (int i = 0; i <= n + 1; ++i)
    for (int j = 1; j <= n; ++j) b[i][j] = c[i][j] = -inf;
  for (int i = n; i >= 1; --i) {
    cnt = f = s = 0;
    for (int j = 1; j <= n; ++j) {
      if (i - j + 1 == 0) break;
      if (a[i - j + 1] == 2) ++cnt;
      if (a[i - j + 1] == 2)
        ++s;
      else
        ++f;
      b[i - j + 1][j] = max(b[i - j + 2][j], f - s);
      if (i + j <= n) ans = max(ans, pr[i - j] + cnt + sf[i + 1] + b[i + 1][j]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    cnt = f = s = 0;
    for (int j = 1; j <= n; ++j) {
      if (i + j - 1 == n + 1) break;
      if (a[i + j - 1] == 1) ++cnt;
      if (a[i + j - 1] == 1)
        ++f;
      else
        ++s;
      c[i][j] = max(c[i - 1][j], s - f);
      if (i - j >= 1) ans = max(ans, sf[i + j] + pr[i - 1] + cnt + c[i - j][j]);
    }
  }
  cout << ans;
  return 0;
}

```