#include <iostream>
using namespace std;

int main() {

	int n, t[52], f;
	char s[52], suit[4] = { 'S','H','C','D' };

	cin >> n;

	for (int i = 0; i < n; i++) {
		cin >> s[i] >> t[i];
	}

	for (int i = 0; i < 4; i++) {
		for (int j = 1; j <= 13; j++) {
			f = 1;
			for (int k = 0; k < n; k++) {
				if (s[k] == suit[i] && t[k] == j) {
					f = 0;
				}
			}
			if (f == 1) {
				cout << suit[i] << " " << j << "\n";
			}
		}
	}
	

	return 0;
}
### Prompt

Please formulate 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, t[52], f;
	char s[52], suit[4] = { 'S','H','C','D' };

	cin >> n;

	for (int i = 0; i < n; i++) {
		cin >> s[i] >> t[i];
	}

	for (int i = 0; i < 4; i++) {
		for (int j = 1; j <= 13; j++) {
			f = 1;
			for (int k = 0; k < n; k++) {
				if (s[k] == suit[i] && t[k] == j) {
					f = 0;
				}
			}
			if (f == 1) {
				cout << suit[i] << " " << j << "\n";
			}
		}
	}
	

	return 0;
}
```