#include <bits/stdc++.h>
using namespace std;
int a[2010];
int dp[2010][2][2010];
int psa[2][2010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    psa[0][i + 1] = psa[0][i];
    psa[1][i + 1] = psa[1][i];
    psa[a[i] - 1][i + 1]++;
  }
  for (int st = n - 1; st >= 0; st--) {
    for (int i = st; i >= 0; i--) {
      dp[st][0][i] = dp[st][0][i + 1] + (a[i] == 1 ? 1 : 0);
      dp[st][1][i] = dp[st][1][i + 1];
      if (a[i] == 2)
        dp[st][1][i] =
            max(dp[st][1][i], max(dp[st][0][i + 1] + 1, dp[st][1][i + 1] + 1));
    }
  }
  int maxi = 0;
  for (int i = 0; i <= n; i++)
    maxi = max(maxi, psa[0][i] + psa[1][n] - psa[1][i]);
  for (int l = 0; l < n; l++)
    for (int r = l; r < n; r++)
      maxi = max(maxi, psa[0][l] + dp[r][1][l] + psa[1][n] - psa[1][r + 1]);
  printf("%d", maxi);
}

### Prompt

Please create a solution in Cpp 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 a[2010];
int dp[2010][2][2010];
int psa[2][2010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    psa[0][i + 1] = psa[0][i];
    psa[1][i + 1] = psa[1][i];
    psa[a[i] - 1][i + 1]++;
  }
  for (int st = n - 1; st >= 0; st--) {
    for (int i = st; i >= 0; i--) {
      dp[st][0][i] = dp[st][0][i + 1] + (a[i] == 1 ? 1 : 0);
      dp[st][1][i] = dp[st][1][i + 1];
      if (a[i] == 2)
        dp[st][1][i] =
            max(dp[st][1][i], max(dp[st][0][i + 1] + 1, dp[st][1][i + 1] + 1));
    }
  }
  int maxi = 0;
  for (int i = 0; i <= n; i++)
    maxi = max(maxi, psa[0][i] + psa[1][n] - psa[1][i]);
  for (int l = 0; l < n; l++)
    for (int r = l; r < n; r++)
      maxi = max(maxi, psa[0][l] + dp[r][1][l] + psa[1][n] - psa[1][r + 1]);
  printf("%d", maxi);
}

```