#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i = 0; i < (n); ++i)
#define ll long long
#define P pair<ll,ll>
#define all(v) v.begin(),v.end()
const ll mod = 1e9+7;
const ll INF = 1e18;
const double pi = acos(-1.0);

int main(void)
{
   
    ll n; cin>>n;
    map<pair<char,ll>,bool> mp;
    rep(i,n){
        char c; ll a;
        cin>>c>>a;
        mp[make_pair(c,a)]=true;
    }
    char im[]={'S','H','C','D'};
    map<char,ll> ans;
    rep(i,4)rep(j,13){
        if(!mp[{im[i],j+1}]) cout<<im[i]<<" "<<j+1<<endl;
    }
    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 <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i = 0; i < (n); ++i)
#define ll long long
#define P pair<ll,ll>
#define all(v) v.begin(),v.end()
const ll mod = 1e9+7;
const ll INF = 1e18;
const double pi = acos(-1.0);

int main(void)
{
   
    ll n; cin>>n;
    map<pair<char,ll>,bool> mp;
    rep(i,n){
        char c; ll a;
        cin>>c>>a;
        mp[make_pair(c,a)]=true;
    }
    char im[]={'S','H','C','D'};
    map<char,ll> ans;
    rep(i,4)rep(j,13){
        if(!mp[{im[i],j+1}]) cout<<im[i]<<" "<<j+1<<endl;
    }
    return 0;
}

```