#include <iostream>
using namespace std;

int main()
{
  int n;
  cin >> n;
  char c[n];
  int a[n];
  for(int i=0;i<n;i++)
    {
      cin >> c[i] >> a[i];
    }
 
  char x[52];
  int y[52];

  for(int i=0;i<13;i++)
    {
      x[i]='S';     y[i]=i+1;
      x[i+13*1]='H';y[i+13*1]=i+1;
      x[i+13*2]='C';y[i+13*2]=i+1;
      x[i+13*3]='D';y[i+13*3]=i+1;
    }

  for(int i=0;i<52;i++)
    {
      for(int j=0;j<n;j++)
	{
	  if(x[i]==c[j]&&y[i]==a[j])goto out;
	 

	}
      cout << x[i] << " " << y[i] << endl;
    out:
      ;
    }

  return 0;
}

    
### Prompt

Your challenge is to write a Cpp solution to the following problem:

Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers).

The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.

Note

解説



Input

In the first line, the number of cards n (n ≤ 52) is given.

In the following n lines, data of the n cards are given. Each card is given by a pair of a character and an integer which represent its suit and rank respectively. A suit is represented by 'S', 'H', 'C' and 'D' for spades, hearts, clubs and diamonds respectively. A rank is represented by an integer from 1 to 13.

Output

Print the missing cards. The same as the input format, each card should be printed with a character and an integer separated by a space character in a line. Arrange the missing cards in the following priorities:

* Print cards of spades, hearts, clubs and diamonds in this order.
* If the suits are equal, print cards with lower ranks first.

Example

Input

47
S 10
S 11
S 12
S 13
H 1
H 2
S 6
S 7
S 8
S 9
H 6
H 8
H 9
H 10
H 11
H 4
H 5
S 2
S 3
S 4
S 5
H 12
H 13
C 1
C 2
D 1
D 2
D 3
D 4
D 5
D 6
D 7
C 3
C 4
C 5
C 6
C 7
C 8
C 9
C 10
C 11
C 13
D 9
D 10
D 11
D 12
D 13


Output

S 1
H 3
H 7
C 12
D 8

### Response

```cpp
#include <iostream>
using namespace std;

int main()
{
  int n;
  cin >> n;
  char c[n];
  int a[n];
  for(int i=0;i<n;i++)
    {
      cin >> c[i] >> a[i];
    }
 
  char x[52];
  int y[52];

  for(int i=0;i<13;i++)
    {
      x[i]='S';     y[i]=i+1;
      x[i+13*1]='H';y[i+13*1]=i+1;
      x[i+13*2]='C';y[i+13*2]=i+1;
      x[i+13*3]='D';y[i+13*3]=i+1;
    }

  for(int i=0;i<52;i++)
    {
      for(int j=0;j<n;j++)
	{
	  if(x[i]==c[j]&&y[i]==a[j])goto out;
	 

	}
      cout << x[i] << " " << y[i] << endl;
    out:
      ;
    }

  return 0;
}

    
```