// DP solution: try all subsets of tools still available

#include <iostream>
#include <algorithm>

using namespace std;

const int MAX = 300000;
int T;
int task[MAX];

// what tools are still available, current tool, task
int memo[8][3][MAX+1];

int solve(int tools, int curr_tool, int task_i)
{
  if (task_i >= T || !tools) {
    return 0;
  }
  
  int &ans = memo[tools][curr_tool][task_i];
  if (ans >= 0) {
    return ans;
  }

  ans = 0;
  
  // try doing this task
  if (task[task_i] == curr_tool) {
    ans = max(ans, solve(tools, curr_tool, task_i+1) + 1);
  }

  // try skipping
  ans = max(ans, solve(tools, curr_tool, task_i+1));

  // try dropping the tool
  int new_tools = tools - (1 << curr_tool);
  for (int i = 0; i < 3; i++) {
    if (new_tools & (1 << i)) {
      ans = max(ans, solve(new_tools, i, task_i));
    }
  }

  return ans;
}

int main()
{
  cin >> T;
  for (int i = 0; i < MAX; i++) {
    cin >> task[i];
  }

  for (int i = 0; i < 8; i++) {
    for (int j = 0; j < 3; j++) {
      fill(memo[i][j], memo[i][j]+MAX+1, -1);
    }
  }

  int ans = 0;
  for (int i = 0; i < 3; i++) {
    ans = max(ans, solve(7, i, 0));
  }
  
  cout << ans << endl;
  
  return 0;
}