#include <bits/stdc++.h>
using namespace std;
const int SI = 1e6 + 1000;
int N;
int sum2[2005], sum1[2005];
int dp[2005][2005][3];
int a[2005];
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
    if (a[i] == 1) {
      sum1[i] = sum1[i - 1] + 1;
      sum2[i] = sum2[i - 1];
    } else {
      sum1[i] = sum1[i - 1];
      sum2[i] = sum2[i - 1] + 1;
    }
  }
  if (sum1[N] == N || sum2[N] == N) {
    cout << N;
    return 0;
  }
  for (int i = 1; i <= N; i++) {
    if (sum1[i] + sum2[N] - sum2[i] == N) {
      cout << N;
      return 0;
    }
  }
  for (int i = 1; i <= N; i++)
    for (int j = i; j <= N; j++) {
      if (a[j] == 1) {
        dp[i][j][1] = ((dp[i][j - 1][2] + 1) > (dp[i][j - 1][1] + 1)
                           ? (dp[i][j - 1][2] + 1)
                           : (dp[i][j - 1][1] + 1));
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  int ans = 0;
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++) {
      if (a[i] == 2 && a[j] == 1) {
        ans = ((ans) > (sum1[i] + sum2[N] - sum2[j] + dp[i][j][1])
                   ? (ans)
                   : (sum1[i] + sum2[N] - sum2[j] + dp[i][j][1]));
      }
    }
  cout << ans;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int SI = 1e6 + 1000;
int N;
int sum2[2005], sum1[2005];
int dp[2005][2005][3];
int a[2005];
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
    if (a[i] == 1) {
      sum1[i] = sum1[i - 1] + 1;
      sum2[i] = sum2[i - 1];
    } else {
      sum1[i] = sum1[i - 1];
      sum2[i] = sum2[i - 1] + 1;
    }
  }
  if (sum1[N] == N || sum2[N] == N) {
    cout << N;
    return 0;
  }
  for (int i = 1; i <= N; i++) {
    if (sum1[i] + sum2[N] - sum2[i] == N) {
      cout << N;
      return 0;
    }
  }
  for (int i = 1; i <= N; i++)
    for (int j = i; j <= N; j++) {
      if (a[j] == 1) {
        dp[i][j][1] = ((dp[i][j - 1][2] + 1) > (dp[i][j - 1][1] + 1)
                           ? (dp[i][j - 1][2] + 1)
                           : (dp[i][j - 1][1] + 1));
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  int ans = 0;
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++) {
      if (a[i] == 2 && a[j] == 1) {
        ans = ((ans) > (sum1[i] + sum2[N] - sum2[j] + dp[i][j][1])
                   ? (ans)
                   : (sum1[i] + sum2[N] - sum2[j] + dp[i][j][1]));
      }
    }
  cout << ans;
  return 0;
}

```