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)

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

int main() {
  int n; cin >> n;
  vector<int> v(3);
  vector<char> ans(n);
  int ind=0;
  cin >> v[0] >> v[1] >> v[2];
  int t=v[0]+v[1]+v[2];
  bool ok=1;
  while (n--) {
    string s; cin >> s;
    if (!ok) continue;
    if (!v[s[0]-'A'] && !v[s[1]-'A']) {
      ok=0;
      continue;
    }
    if (!v[s[0]-'A']) v[s[0]-'A']++, v[s[1]-'A']--, ans[ind++]=s[0];
    else if (!v[s[1]-'A']) v[s[1]-'A']++, v[s[0]-'A']--, ans[ind++]=s[1];
    else {
      if (t == 2) {
        string c="#";
        if (n--) {
          cin >> c;
        }
        if (c != "#") {
          if (s[0] == c[0] || s[0] == c[1])
            v[s[0]-'A']++, v[s[1]-'A']--, ans[ind++]=s[0];
          else
            v[s[1]-'A']++, v[s[0]-'A']--, ans[ind++]=s[1];
        }
        else 
          v[s[0]-'A']++, v[s[1]-'A']--, ans[ind++]=s[0];
      }
      else {
        if (v[s[1]-'A'] == t-1)
          v[s[0]-'A']++, v[s[1]-'A']--, ans[ind++]=s[0];
        else
          v[s[1]-'A']++, v[s[0]-'A']--, ans[ind++]=s[1];
      }
    }
  }
  if (n == 0 || !ok) cout << "No";
  else {
    cout << "Yes\n";
    for (char c : ans)
      cout << c << "\n";
  }
  return 0;
}