#include <bits/stdc++.h>
using namespace std;
vector<int> cnt;
const int Inf = 1e9;
int countOne(int l, int r) { return r - l - cnt[r] + cnt[l]; }
int countTwo(int l, int r) { return r - l - countOne(l, r); }
int countSecond(int l, int r) { return countTwo(l, r) - countOne(l, r); }
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  cnt.resize(n + 1);
  cnt[0] = 0;
  for (int i = 1; i <= n; ++i) {
    cnt[i] = cnt[i - 1] + (a[i - 1] == 2);
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, countOne(0, i) + countTwo(i, n));
  }
  for (int l = 0; l < n; ++l) {
    int cur = -Inf;
    for (int r = l; r < n; ++r) {
      cur = max(cur, countSecond(l, r + 1));
      ans = max(ans,
                countOne(l, r + 1) + cur + countOne(0, l) + countTwo(r + 1, n));
    }
  }
  cout << ans << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<int> cnt;
const int Inf = 1e9;
int countOne(int l, int r) { return r - l - cnt[r] + cnt[l]; }
int countTwo(int l, int r) { return r - l - countOne(l, r); }
int countSecond(int l, int r) { return countTwo(l, r) - countOne(l, r); }
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  cnt.resize(n + 1);
  cnt[0] = 0;
  for (int i = 1; i <= n; ++i) {
    cnt[i] = cnt[i - 1] + (a[i - 1] == 2);
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, countOne(0, i) + countTwo(i, n));
  }
  for (int l = 0; l < n; ++l) {
    int cur = -Inf;
    for (int r = l; r < n; ++r) {
      cur = max(cur, countSecond(l, r + 1));
      ans = max(ans,
                countOne(l, r + 1) + cur + countOne(0, l) + countTwo(r + 1, n));
    }
  }
  cout << ans << endl;
}

```