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)

/* Author: $%U%$
 * Time: $%Y%$-$%M%$-$%D%$ $%h%$:$%m%$:$%s%$
**/
#include <bits/stdc++.h>
using namespace std;
#define fore(i, a, b) for (int i = a, to = b; i < to; i++)
#define all(v) v.begin(), v.end()
#define SZ(v) (int)v.size()
#define pb push_back
typedef long long Long;
typedef pair<int, int> Pair;
int comun(string a, string b) {
  for (char x = 'A'; x <= 'C'; x++) {
    int cnt = 0;
    fore(i, 0, 2) if (x == a[i]) cnt++;
    fore(i, 0, 2) if (x == b[i]) cnt++;
    if (cnt == 2) return x - 'A';
  }
  return -1;
}
vector<Long> val;
string res;
bool fil(deque<string> &v) {
  fore(i, 0, SZ(v)) {
    int p1 = v[i][0] - 'A';
    int p2 = v[i][1] - 'A';
    if (val[p1] == 0) {
      res.pb(char(p1 + 'A'));
      val[p1]++;
      val[p2]--;
    } else if (val[p2] == 0) {
      res.pb(char(p2 + 'A'));
      val[p2]++;
      val[p1]--;
    } else if (max(val[p1], val[p2]) > 1) {
      if (val[p1] > val[p2]) {
        res.pb(char(p2 + 'A'));
        val[p2]++;
        val[p1]--;
      } else {
        res.pb(char(p1 + 'A'));
        val[p1]++;
        val[p2]--;
      }
    } else if (i + 1 < SZ(v)) {
      //hay uno extra
      string extra = v[i + 1];
      if (extra == v[i]) {
        res.pb(char(p1 + 'A'));
        res.pb(char(p2 + 'A'));
        i++;
      } else {
        int com = comun(v[i], extra);
        int ini = com ^ p1 ^ p2;
        res.pb(char(com + 'A'));
        val[ini]--;
        val[com]++;
      }
    } else {
      res.pb(char(p1 + 'A'));
      val[p1]++;
      val[p2]--;
    }
    fore(_, 0, 3) if (val[_] < 0) return false;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  val.resize(3);
  fore(i, 0, 3) cin >> val[i];
  deque<string> v(n);
  fore(i, 0, n) cin >> v[i];
  int cnt = 0;
  fore(i, 0, n) if (val[i] > 0) cnt++;
  if (cnt == 0) {
    cout << "No\n";
  } else {
    if (fil(v)) {
      cout << "Yes\n";
      for (char x : res) cout << x << '\n';
    } else {
      cout << "No\n";
    }
  }

  return 0;
}