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: Accepted)

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*/

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

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second

template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class T>
istream& operator>>(istream& is, vector<T>& xs){
  for(auto& x: xs)
	is >> x;
  return is;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
ostream& operator<<(ostream& os, const vector<T>& xs){
  for(unsigned int i=0;i<xs.size();++i)
	os << (i?" ":"") << xs[i];
  return os;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}
void debug(istringstream&){}
template <char sep=',', class Head, class... Tail>
void debug(istringstream& iss, Head&& head, Tail&&... tail)
{
  string name;
  getline(iss, name, ',');
  cout << sep << name << "=" << head;
  debug(iss, forward<Tail>(tail)...);
}

#ifdef PYONPOI
#define DEBUG(...) \
  do{ \
    istringstream ss(#__VA_ARGS__); \
    debug<' '>(ss, __VA_ARGS__); \
    cout<<endl; \
  }while(0)
#else
#define DEBUG
#endif
void init_io(int argc, char* argv[]){
  cin.tie(0);
  ios_base::sync_with_stdio(false);
#ifdef PYONPOI
  if(argc > 1) {
    ifstream *ifs = new ifstream();
    ifs->open(argv[1]);
    cin.rdbuf(ifs->rdbuf());
  }
#endif
}


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

int main(int argc, char* argv[]){
  init_io(argc, argv);

  int N;
  VL xs(3);
  cin >> N >> xs;

  VS vs(N);
  cin >> vs;
  VI type(N);
  REP(i,N) type[i] = (vs[i] == "AB" ? 2 : vs[i] == "AC" ? 1 : 0);

  VI ntype(N, -1);
  for(int i=N-2;i>=0;--i){
    if(type[i] != type[i+1]) ntype[i] = type[i+1];
    else ntype[i] = ntype[i+1];
  }

  bool ok = true;
  VI ans(N);
  LL s = xs[0] + xs[1] + xs[2];
  if(s == 0) {
    ok = false;
  }
  else {
    REP(i,N){
      int a = (type[i] + 1) % 3;
      int b = 3 - type[i] - a;

      int add = -1, del = -1;
      DEBUG(a,b,type[i], xs[a],xs[b]);
      if(xs[a] > xs[b]){
        add = b;
        del = a;
      }
      else if (xs[a] < xs[b]){
        add = a;
        del = b;
      }
      else {
        if(xs[a] > 0) {
          int nt = ntype[i];
          if(nt != -1) {
            if(a == nt){
              add = b;
              del = a;
            }
            else {
              add = a;
              del = b;
            }
          }
          else {
            add = a;
            del = b;
          }
        }
      }

      if(add < 0){
        ok = false;
        break;
      }
      xs[add]++;
      xs[del]--;
      ans[i] = add;
    }
  }

  if(ok){
    cout << "Yes" << endl;
    for(auto i : ans) cout << ("ABC"[i]) << endl;
  }
  else cout << "No" << endl;

  return 0;
}