#include <bits/stdc++.h>
using namespace std;
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long dp[4][100003];
int main() {
  long long n;
  long long a[100003];
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    if (i == 0) {
      if (a[i] == 1) {
        dp[0][0]++;
        dp[2][0]++;
      } else {
        dp[1][0]++;
        dp[3][0]++;
      }
      continue;
    }
    if (a[i] == 1) {
      dp[0][i] = dp[0][i - 1] + 1;
      dp[2][i] = max(dp[2][i - 1], dp[1][i - 1]) + 1;
      dp[1][i] = dp[1][i - 1];
      dp[3][i] = dp[3][i - 1];
    } else {
      dp[1][i] = max(dp[1][i - 1], dp[0][i - 1]) + 1;
      dp[3][i] = max(dp[2][i - 1], dp[3][i - 1]) + 1;
      dp[0][i] = dp[0][i - 1];
      dp[2][i] = dp[2][i - 1];
    }
  }
  cout << max(dp[0][n - 1], max(dp[1][n - 1], max(dp[2][n - 1], dp[3][n - 1])))
       << "\n";
}

### 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;
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long dp[4][100003];
int main() {
  long long n;
  long long a[100003];
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    if (i == 0) {
      if (a[i] == 1) {
        dp[0][0]++;
        dp[2][0]++;
      } else {
        dp[1][0]++;
        dp[3][0]++;
      }
      continue;
    }
    if (a[i] == 1) {
      dp[0][i] = dp[0][i - 1] + 1;
      dp[2][i] = max(dp[2][i - 1], dp[1][i - 1]) + 1;
      dp[1][i] = dp[1][i - 1];
      dp[3][i] = dp[3][i - 1];
    } else {
      dp[1][i] = max(dp[1][i - 1], dp[0][i - 1]) + 1;
      dp[3][i] = max(dp[2][i - 1], dp[3][i - 1]) + 1;
      dp[0][i] = dp[0][i - 1];
      dp[2][i] = dp[2][i - 1];
    }
  }
  cout << max(dp[0][n - 1], max(dp[1][n - 1], max(dp[2][n - 1], dp[3][n - 1])))
       << "\n";
}

```