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<iostream>
#include<string>
using namespace std;
int N_MAX=100000;
int main()
{
  int n,i,j,k,jj,kk;
  int a[3];
  string s,sb;
  char res[N_MAX];
  cin >> n >> a[0] >> a[1] >> a[2];
  cin >> sb;
  for(i=0;i<n;i++){
    if(i<n-1)
      cin >> s;
    j=(int)(sb[0]-'A');
    k=(int)(sb[1]-'A');
    jj=(int)(s[0]-'A');
    kk=(int)(s[1]-'A');
    if(a[j]==0&&a[k]==0){
      cout << "No" << endl;
      return 0;
    }
    if(a[j]>a[k]
       ||a[j]==a[k]&&(i==n-1||k==jj&&a[kk]==0||k==kk&&a[jj]==0)){
      a[j]--;
      a[k]++;
      res[i]=(char)('A'+k);
    }else{
      a[j]++;
      a[k]--;
      res[i]=(char)('A'+j);
    }
    sb=s;
  }
  cout << "Yes" << endl;
  for(i=0;i<n;i++)
    cout << res[i] << endl;
  return 0;
}