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 <vector>
#include <bitset>
#include <utility>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
 
using namespace std;
typedef long long ll;
 
int main(){
  int N,A,B,C;
  cin >> N >> A >> B >> C;
  vector<int> s(N);
  string si;
  for(int i=0;i<N;i++){
    cin >> si;
    if(si=="AB") s[i]=0;
    else if(si=="BC") s[i]=1;
    else s[i]=2;
  }

  vector<string> abc={"A","B","C"};
  vector<int> cur(3);
  cur={A,B,C};
  vector<int> ans(N);
  int l,m;
  for(int i=0;i<N;i++){
    l=s[i]; m=(l+1)%3;
    if(cur[l]<cur[m]){
      ans[i]=l;
      cur[l]++; cur[m]--;
    }else if(cur[l]==1 && cur[m]==1 && i<N-1){
      if(m!=s[i+1] && m!=(s[i+1]+1)%3){
        ans[i]=l;
        cur[l]++; cur[m]--;
      }else{
        ans[i]=m;
        cur[m]++; cur[l]--;
      }
    }else{
      cur[m]++; cur[l]--;
      ans[i]=m;
    }
    if(cur[0]<0 || cur[1]<0 || cur[2]<0){
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  for(int i=0;i<N;i++){
    cout << abc[ans[i]] << endl;
  }

  return 0;
}