Problem: There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

* 1 \leq N \leq 10^5
* 0 \leq A,B,C \leq 10^9
* N, A, B, C are integers.
* s_i is `AB`, `AC`, or `BC`.

Input

Input is given from Standard Input in the following format:


N A B C
s_1
s_2
:
s_N


Output

If it is possible to make N choices under the condition, print `Yes`; otherwise, print `No`.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (`A`, `B`, or `C`) to which you add 1 in the i-th choice.

Examples

Input

2 1 3 0
AB
AC


Output

Yes
A
C


Input

3 1 0 0
AB
BC
AB


Output

No


Input

1 0 9 0
AC


Output

No


Input

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB


Output

Yes
C
B
B
C
C
B
A
A

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>

using namespace std;

const int MAXN=1e5+5;
int N,ABC[3];
char buf[4];

int main(){
    scanf("%d%d%d%d",&N,ABC,ABC+1,ABC+2);

    vector<char> ans;
    bool inDanger=false;
    for(int i=0;i<N;i++){
        if (!inDanger)
            scanf("%s",buf);
        else 
            inDanger=false;
        int fst=buf[0]-'A',snd=buf[1]-'A';
        if (ABC[fst]==0&&ABC[snd]==0){
            puts("No");
            return 0;
        } 

        if (ABC[fst]==1&&ABC[snd]==1){
            scanf("%s",buf);
            int nxtFst=buf[0]-'A',nxtSnd=buf[1]-'A';
            if (fst==nxtFst||fst==nxtSnd)
                ans.push_back(fst+'A'),ABC[fst]++,ABC[snd]--;
            else 
                ans.push_back(snd+'A'),ABC[fst]--,ABC[snd]++;

            inDanger=true;
            continue;
        }  

        if (ABC[fst]>ABC[snd])
            ans.push_back(snd+'A'),ABC[fst]--,ABC[snd]++;
        else 
            ans.push_back(fst+'A'),ABC[fst]++,ABC[snd]--;
    }

    puts("Yes");
    for(auto &c:ans) printf("%c\n",c);

    return 0;
}