#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e3 + 11;
long long n;
long long a[maxn];
long long par[3][maxn];
long long dc[maxn][maxn];
long long query(long long *a, long long l, long long r) {
  if (l < 0 || r < l) return 0;
  return a[r] - ((l >= 1) ? a[l - 1] : 0);
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long nu = 1; nu <= 2; nu++) {
    for (long long i = 0; i < n; i++) {
      par[nu][i] = (a[i] == nu) + ((i >= 1) ? par[nu][i - 1] : 0);
    }
  }
  for (long long i = 0; i < n; i++) {
    dc[i][i] = 1;
    for (long long j = i + 1; j < n; j++) {
      dc[i][j] = query(par[1], i, j);
      dc[i][j] = max(dc[i][j], query(par[2], i, j));
      if (a[j] == 1)
        dc[i][j] = max(dc[i][j], dc[i][j - 1] + 1);
      else
        dc[i][j] = max(dc[i][j], dc[i][j - 1]);
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++) {
      long long temp = 0;
      temp += query(par[1], 0, i - 1);
      temp += query(par[2], j + 1, n - 1);
      temp += dc[i][j];
      ans = max(ans, temp);
    }
  cout << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 maxn = 2e3 + 11;
long long n;
long long a[maxn];
long long par[3][maxn];
long long dc[maxn][maxn];
long long query(long long *a, long long l, long long r) {
  if (l < 0 || r < l) return 0;
  return a[r] - ((l >= 1) ? a[l - 1] : 0);
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long nu = 1; nu <= 2; nu++) {
    for (long long i = 0; i < n; i++) {
      par[nu][i] = (a[i] == nu) + ((i >= 1) ? par[nu][i - 1] : 0);
    }
  }
  for (long long i = 0; i < n; i++) {
    dc[i][i] = 1;
    for (long long j = i + 1; j < n; j++) {
      dc[i][j] = query(par[1], i, j);
      dc[i][j] = max(dc[i][j], query(par[2], i, j));
      if (a[j] == 1)
        dc[i][j] = max(dc[i][j], dc[i][j - 1] + 1);
      else
        dc[i][j] = max(dc[i][j], dc[i][j - 1]);
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++) {
      long long temp = 0;
      temp += query(par[1], 0, i - 1);
      temp += query(par[2], j + 1, n - 1);
      temp += dc[i][j];
      ans = max(ans, temp);
    }
  cout << ans << endl;
  return 0;
}

```