#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 10;
int n, a[MAXN], cnt1[MAXN], cnt2[MAXN];
int get(int i, int j, int cnt[MAXN]) { return cnt[j] - cnt[i]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  cnt1[0] = cnt2[0] = 0;
  for (int i = 0; i < int(n); i++) {
    cnt1[i + 1] = cnt1[i] + (a[i] == 1);
    cnt2[i + 1] = cnt2[i] + (a[i] == 2);
  }
  int best = 0;
  for (int k = 0; k < int(n); k++) {
    int best21 = k + 1, best12 = k + 1;
    for (int i = (int)(k + 1 - 1); i >= int(0); i--)
      if (get(i, k + 1, cnt2) - get(i, k + 1, cnt1) >
          get(best21, k + 1, cnt2) - get(best21, k + 1, cnt1))
        best21 = i;
    for (int i = (int)(k + 1); i < (int)(n + 1); i++)
      if (get(k + 1, i, cnt1) - get(k + 1, i, cnt2) >
          get(k + 1, best12, cnt1) - get(k + 1, best12, cnt2))
        best12 = i;
    best = max(best, get(0, best21, cnt1) + get(best21, k + 1, cnt2) +
                         get(k + 1, best12, cnt1) + get(best12, n, cnt2));
  }
  cout << best << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAXN = 2e3 + 10;
int n, a[MAXN], cnt1[MAXN], cnt2[MAXN];
int get(int i, int j, int cnt[MAXN]) { return cnt[j] - cnt[i]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  cnt1[0] = cnt2[0] = 0;
  for (int i = 0; i < int(n); i++) {
    cnt1[i + 1] = cnt1[i] + (a[i] == 1);
    cnt2[i + 1] = cnt2[i] + (a[i] == 2);
  }
  int best = 0;
  for (int k = 0; k < int(n); k++) {
    int best21 = k + 1, best12 = k + 1;
    for (int i = (int)(k + 1 - 1); i >= int(0); i--)
      if (get(i, k + 1, cnt2) - get(i, k + 1, cnt1) >
          get(best21, k + 1, cnt2) - get(best21, k + 1, cnt1))
        best21 = i;
    for (int i = (int)(k + 1); i < (int)(n + 1); i++)
      if (get(k + 1, i, cnt1) - get(k + 1, i, cnt2) >
          get(k + 1, best12, cnt1) - get(k + 1, best12, cnt2))
        best12 = i;
    best = max(best, get(0, best21, cnt1) + get(best21, k + 1, cnt2) +
                         get(k + 1, best12, cnt1) + get(best12, n, cnt2));
  }
  cout << best << '\n';
  return 0;
}

```