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 <array>
#include <functional>
#include <algorithm>
using namespace std;

int main() {
  int64_t n;
  array<int64_t, 3> abc;
  cin >> n >> abc[0] >> abc[1] >> abc[2];
  string out, s[n];
  for (int i = 0; cin >> s[i]; ++i) {}
  function<bool(int)> solve = [&](int i) {
    if (i == n) {
      return true;
    }
    --abc[s[i].front() - 'A'];
    ++abc[s[i].back() - 'A'];
    if (abc[s[i].front() - 'A'] >= 0 && solve(i + 1)) {
      out.push_back(s[i].back());
      out.push_back('\n');
      return true;
    }
    abc[s[i].front() - 'A'] += 2;
    abc[s[i].back() - 'A'] -= 2;
    if (abc[s[i].back() - 'A'] >= 0 && solve(i + 1)) {
      out.push_back(s[i].front());
      out.push_back('\n');
      return true;
    }
    --abc[s[i].front() - 'A'];
    ++abc[s[i].back() - 'A'];
    return false;
  };
  if (solve(0)) {
    reverse(begin(out), end(out));
    cout << "Yes" << out;
  } else {
    cout << "No";
  }
  return 0;
}