#include <bits/stdc++.h>
using namespace std;
const long long logN = 20;
const long long N = 100005;
const long long M = 1000000007;
const long long INF = 1e12;
const long long dx[4] = {1, 0, -1, 0};
const long long dy[4] = {0, 1, 0, -1};
long long A[2005];
long long dp[2005][2005];
long long beg[2005], la[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) cin >> A[i];
  for (long long i = 1; i < n + 1; i++) {
    long long l1 = 0, l2 = 0;
    for (long long j = i; j < n + 1; j++) {
      if (A[j] == 1) {
        l1 = max(l1, l2) + 1;
      } else {
        l2++;
      }
      dp[i][j] = max(l1, l2);
    }
  }
  long long cnt = 0;
  for (long long i = 1; i < n + 1; i++) {
    if (A[i] == 1) cnt++;
    beg[i] = cnt;
  }
  cnt = 0;
  for (long long i = n; i >= 1; i--) {
    if (A[i] == 2) cnt++;
    la[i] = cnt;
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++)
    for (long long j = i; j < n + 1; j++) {
      ans = max(ans, beg[i - 1] + dp[i][j] + la[j + 1]);
    }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
const long long logN = 20;
const long long N = 100005;
const long long M = 1000000007;
const long long INF = 1e12;
const long long dx[4] = {1, 0, -1, 0};
const long long dy[4] = {0, 1, 0, -1};
long long A[2005];
long long dp[2005][2005];
long long beg[2005], la[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) cin >> A[i];
  for (long long i = 1; i < n + 1; i++) {
    long long l1 = 0, l2 = 0;
    for (long long j = i; j < n + 1; j++) {
      if (A[j] == 1) {
        l1 = max(l1, l2) + 1;
      } else {
        l2++;
      }
      dp[i][j] = max(l1, l2);
    }
  }
  long long cnt = 0;
  for (long long i = 1; i < n + 1; i++) {
    if (A[i] == 1) cnt++;
    beg[i] = cnt;
  }
  cnt = 0;
  for (long long i = n; i >= 1; i--) {
    if (A[i] == 2) cnt++;
    la[i] = cnt;
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++)
    for (long long j = i; j < n + 1; j++) {
      ans = max(ans, beg[i - 1] + dp[i][j] + la[j + 1]);
    }
  cout << ans << "\n";
  return 0;
}

```