#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, i, j, k = 0;
  cin >> n;
  long long ar[n];
  for (i = 0; i < n; i++) cin >> ar[i];
  long long one[n + 1], two[n], dp[n][n];
  for (i = 0; i < n; i++) {
    one[i] = two[i] = 0;
  }
  for (i = 1; i < n; i++) {
    if (ar[i - 1] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  }
  if (ar[n - 1] == 1)
    one[n] = one[n - 1] + 1;
  else
    one[n] = one[n - 1];
  for (i = n - 2; i >= 0; i--) {
    if (ar[i + 1] == 2)
      two[i] = two[i + 1] + 1;
    else
      two[i] = two[i + 1];
  }
  for (i = 0; i < n; i++) dp[i][i] = 1;
  for (j = n - 1; j > 0; j--) {
    for (i = j - 1; i >= 0; i--) {
      dp[i][j] = dp[i + 1][j];
      if (ar[i] == 2 || dp[i + 1][j] == one[j + 1] - one[i + 1])
        dp[i][j] = dp[i + 1][j] + 1;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      k = fmax(k, one[i] + two[j] + dp[i][j]);
    }
  }
  cout << k;
  return 0;
}

### Prompt

Your task is to create 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() {
  long long n, i, j, k = 0;
  cin >> n;
  long long ar[n];
  for (i = 0; i < n; i++) cin >> ar[i];
  long long one[n + 1], two[n], dp[n][n];
  for (i = 0; i < n; i++) {
    one[i] = two[i] = 0;
  }
  for (i = 1; i < n; i++) {
    if (ar[i - 1] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  }
  if (ar[n - 1] == 1)
    one[n] = one[n - 1] + 1;
  else
    one[n] = one[n - 1];
  for (i = n - 2; i >= 0; i--) {
    if (ar[i + 1] == 2)
      two[i] = two[i + 1] + 1;
    else
      two[i] = two[i + 1];
  }
  for (i = 0; i < n; i++) dp[i][i] = 1;
  for (j = n - 1; j > 0; j--) {
    for (i = j - 1; i >= 0; i--) {
      dp[i][j] = dp[i + 1][j];
      if (ar[i] == 2 || dp[i + 1][j] == one[j + 1] - one[i + 1])
        dp[i][j] = dp[i + 1][j] + 1;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      k = fmax(k, one[i] + two[j] + dp[i][j]);
    }
  }
  cout << k;
  return 0;
}

```