#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int one[n], two[n];
    if (a[0] == 1) {
      one[0] = 1;
      two[0] = 0;
    } else {
      one[0] = 0;
      two[0] = 1;
    }
    for (long long int i = 1; i < n; i++) {
      if (a[i] == 1) {
        one[i] = one[i - 1] + 1;
        two[i] = two[i - 1];
      } else {
        one[i] = one[i - 1];
        two[i] = two[i - 1] + 1;
      }
    }
    long long int ans = 1;
    for (long long int i = 0; i < n; i++) {
      long long int l12 = 0, r12 = 0;
      for (long long int j = 0; j < i; j++) {
        l12 = max(l12, one[j] + two[i - 1] - two[j]);
      }
      if (i) {
        l12 = max(l12, two[i - 1]);
        l12 = max(l12, one[i - 1]);
      }
      for (long long int j = i; j < n; j++) {
        r12 = max(r12, one[j] - (i > 0 ? one[i - 1] : 0) + two[n - 1] - two[j]);
      }
      ans = max(ans, r12 + l12);
    }
    cout << ans << "\n";
  }
  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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int one[n], two[n];
    if (a[0] == 1) {
      one[0] = 1;
      two[0] = 0;
    } else {
      one[0] = 0;
      two[0] = 1;
    }
    for (long long int i = 1; i < n; i++) {
      if (a[i] == 1) {
        one[i] = one[i - 1] + 1;
        two[i] = two[i - 1];
      } else {
        one[i] = one[i - 1];
        two[i] = two[i - 1] + 1;
      }
    }
    long long int ans = 1;
    for (long long int i = 0; i < n; i++) {
      long long int l12 = 0, r12 = 0;
      for (long long int j = 0; j < i; j++) {
        l12 = max(l12, one[j] + two[i - 1] - two[j]);
      }
      if (i) {
        l12 = max(l12, two[i - 1]);
        l12 = max(l12, one[i - 1]);
      }
      for (long long int j = i; j < n; j++) {
        r12 = max(r12, one[j] - (i > 0 ? one[i - 1] : 0) + two[n - 1] - two[j]);
      }
      ans = max(ans, r12 + l12);
    }
    cout << ans << "\n";
  }
  return 0;
}

```