Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

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

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
//#define num 1000000007
#define pi acos(-1.0)
//cout << fixed << setprecision (20);   小数点以下2０桁まで
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'0'+=16;        で大文字に
//'大文字'+=32;    で小文字に
//s[i]-'0'でchar文字→int数に;
//string s = to_string(int数);
//int n = stoi(string文字列)
//実行時間制約2秒では２×10^8回くらいまで計算できる


int main() {
  ll n,a,b,c;
  cin >> n>>a>>b>>c;
  
  vector<string> p(n);
  vector<char> q(n);
  
  if(a+b+c==1) {
    rep(i,n) {
      cin >> p[i];
      if(p[i]=="AB") {
        if(a+b==1) {
          if(a==0) q[i]='A';
          else q[i]='B';
          a=1-a;
          b=1-b;
        }
        else {
        cout << "No" << endl;
        return 0;
        }
      }
      if(p[i]=="BC") {
        if(b+c==1) {
          if(b==0) q[i]='B';
          else q[i]='C';
          c=1-c;
          b=1-b;
        }
        else {
        cout << "No" << endl;
        return 0;
        }
      }
      if(p[i]=="AC") {
        if(a+c==1) {
          if(a==0) q[i]='A';
          else q[i]='C';
          a=1-a;
          c=1-c;
        }
        else {
        cout << "No" << endl;
        return 0;
        }
      }
    }
  }
  else {
    rep(i,n) {
    cin >> p[i];
    }
  }
  
  if((a==0 && b==0 && p[0]=="AB")||(b==0 && c==0 && p[0]=="BC")||
    (c==0 && a==0 && p[0]=="AC")) { 
  cout << "No" << endl;    
  return 0;
  }
  
  cout << "Yes" << endl;
  
  if(a+b+c==1) {
    rep(i,n) {
    cout << q[i] << endl;
    }
  }
  
  else{
    rep(i,n) {
      if(p[i]=="AB") {
        if(a<=b) {
          cout << 'A' << endl;
          a+=1;
          b-=1;
        }
        else {
          cout << 'B' << endl;
          a-=1;
          b+=1;
        }
      }
      if(p[i]=="BC") {
        if(b<=c) {
          cout << 'B' << endl;
          b+=1;
          c-=1;
        }
        else {
          cout << 'C' << endl;
          b-=1;
          c+=1;
        }
      }
      if(p[i]=="AC") {
        if(c<=a) {
          cout << 'C' << endl;
          c+=1;
          a-=1;
        }
        else {
          cout << 'A' << endl;
          c-=1;
          a+=1;
        }
      }
    }
  }
}