#include <iostream>
#include <vector>

using namespace std;
vector<vector<int> > memo;

int gcd(int a, int b){
  while(b != 0){
    int temp = b;
    b = a%b;
    a = temp;
  }
  return a;
}

void Rotate(const vector<int>& orig, vector<int>& copy, int amount){
  for(int i = 0; i < copy.size(); i++){
    copy[i] = orig[(i+amount)%orig.size()];
  }
}


// What is the cost of removing the sequence from x to y? x will be the last 
// element remaining in the range.
int Solve(const vector<int>& sequence, int x, int y){
  if(memo[x][y] !=-1)
    return memo[x][y];
  if(x+1 == y){
    memo[x][y] = gcd(sequence[x], sequence[y]);
    return memo[x][y];
  }
  if(x >=y)
    return 0;
    
  int best = 100*100;
  // the element at position x will be the last thing to go.  
  // find the second to last thing
  for(int i = x+1; i <= y; i++){
    int temp = Solve(sequence, x,i-1) + Solve(sequence, i,y) + gcd(sequence[x], sequence[i]);
    if(temp < best)
      best = temp;
  }
  memo[x][y] = best;
  return best;
}
   
void Print_Vector(const vector<int>& v){
  for(int i = 0; i < v.size(); i++)
    cout << v[i] << " ";
  cout << endl;
}

int main(){
  int cur_case = 1;
  int n;
  cin >> n;

  while(n >0){
    memo.resize(n);
    for(int i = 0; i < n; i++)
      memo[i].resize(n);
    
    vector<int> sequence(n);
    for(int i = 0; i < n; i++)
      cin >> sequence[i];
    int best= 100*100;
    for(int i = 0; i < n; i++){
      vector<int> temp_sequence(n);
      // rotate the sequence so that element i is first.  It's more efficient
      // to do this with clever indec offsets, but this is esier.
      Rotate(sequence, temp_sequence, i);
      // add a copy of the first element to the end to aid wrapping around
      temp_sequence.push_back(temp_sequence[0]); 
      for(int j = 0; j < n; j++)
	for(int k = 0; k < n; k++)
	  memo[j][k] = -1;
      // Print_Vector(temp_sequence);
      int cur = Solve(temp_sequence,0,n-1);
      if(cur < best)
	best = cur;
    }
    cout << "Case " << cur_case <<": " << best << endl;
    cur_case++;
    cin >> n ;
  }
  return 0;
}