#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 5e3 + 3, M = 1e5 + 3, BIG = (long long)1e18 + 3,
                MOD = 1e9 + 7, P = 'z' - 'a' + 1, K = 4;
const double PI = 3.14159265359;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, ans = 0;
  cin >> n;
  vector<int> a(n + 2), pref1(n + 2), pref2(n + 2);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pref1[i] = pref1[i - 1] + (a[i] == 1);
    pref2[i] = pref2[i - 1] + (a[i] == 2);
  }
  for (int i = 0; i <= n; ++i) ans == max(ans, pref1[i] + pref2[n] - pref2[i]);
  for (int L = 1; L <= n; ++L) {
    int ans_now1 = 0, ans_now2 = 0;
    for (int R = L; R <= n; ++R) {
      if (a[R] == 2)
        ++ans_now2;
      else
        ans_now1 = max(ans_now2, ans_now1) + 1;
      ans = max(ans,
                pref1[L - 1] + max(ans_now2, ans_now1) + pref2[n] - pref2[R]);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 5e3 + 3, M = 1e5 + 3, BIG = (long long)1e18 + 3,
                MOD = 1e9 + 7, P = 'z' - 'a' + 1, K = 4;
const double PI = 3.14159265359;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, ans = 0;
  cin >> n;
  vector<int> a(n + 2), pref1(n + 2), pref2(n + 2);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pref1[i] = pref1[i - 1] + (a[i] == 1);
    pref2[i] = pref2[i - 1] + (a[i] == 2);
  }
  for (int i = 0; i <= n; ++i) ans == max(ans, pref1[i] + pref2[n] - pref2[i]);
  for (int L = 1; L <= n; ++L) {
    int ans_now1 = 0, ans_now2 = 0;
    for (int R = L; R <= n; ++R) {
      if (a[R] == 2)
        ++ans_now2;
      else
        ans_now1 = max(ans_now2, ans_now1) + 1;
      ans = max(ans,
                pref1[L - 1] + max(ans_now2, ans_now1) + pref2[n] - pref2[R]);
    }
  }
  cout << ans;
  return 0;
}

```