#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<long long> pref(n + 2, 0), suf(n + 2, 0);
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + (a[i] == 2);
  long long res = 0;
  for (int i = 0; i <= n; i++) {
    res = max(res, pref[i] + suf[i + 1]);
  }
  for (int l = 1; l <= n; l++) {
    long long cnt1 = 0, cnt2 = 0, t = 0;
    for (int r = l; r <= n; r++) {
      if (a[r] == 1)
        cnt1++;
      else
        cnt2++;
      if (cnt2 >= cnt1) {
        t += cnt2;
        cnt1 = 0;
        cnt2 = 0;
      }
      res = max(res, pref[l - 1] + suf[r + 1] + cnt1 + t);
    }
  }
  cout << res;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<long long> pref(n + 2, 0), suf(n + 2, 0);
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + (a[i] == 2);
  long long res = 0;
  for (int i = 0; i <= n; i++) {
    res = max(res, pref[i] + suf[i + 1]);
  }
  for (int l = 1; l <= n; l++) {
    long long cnt1 = 0, cnt2 = 0, t = 0;
    for (int r = l; r <= n; r++) {
      if (a[r] == 1)
        cnt1++;
      else
        cnt2++;
      if (cnt2 >= cnt1) {
        t += cnt2;
        cnt1 = 0;
        cnt2 = 0;
      }
      res = max(res, pref[l - 1] + suf[r + 1] + cnt1 + t);
    }
  }
  cout << res;
  return 0;
}

```