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

#ifdef LOCAL
#include "debug.h"
#else
#define DEBUG(...)
#endif

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, a[3];
  cin >> n >> a[0] >> a[1] >> a[2];
  vector<string> s(n);
  for (auto&& e : s) {
    cin >> e;
  }
  string res;
  for (int t = 0; t < n; ++t) {
    int i = s[t][0] - 'A', j = s[t][1] - 'A';
    if (a[i] == 0 and a[j] == 0) {
      cout << "No\n";
      exit(0);
    }
    if (a[i] == 0) {
      ++a[i], --a[j];
      res += 'A' + i;
    } else if (a[j] == 0) {
      --a[i], ++a[j];
      res += 'A' + j;
    } else if (a[i] == 1 and a[3 ^ i ^ j] == 0 and (t + 1 < n and s[t + 1].find('A' + j) == string::npos)) {
      ++a[i], --a[j];
      res += 'A' + i;
    } else {
      --a[i], ++a[j];
      res += 'A' + j;
    }
  }
  cout << "Yes\n";
  for (auto e : res) {
    cout << e << '\n';
  }
}