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;

int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);

  int n;
  vector<int> val(3);
  cin >> n >> val[0] >> val[1] >> val[2];
  
  vector<pair<int, int>> p(n+1);
  for (int i = 0; i < n; ++i) {
    string t;
    cin >> t;
    p[i].first = t[0] - 'A';
    p[i].second = t[1] - 'A';
  }
   
  string ans;
  auto add = [&](int l, int r) {
    val[l]++;
    val[r]--;
    ans += 'A'+l;
  };

  for (int i = 0; i < n; ++i) {
    int a = p[i].first, b = p[i].second;
    if (!val[a] && !val[b]) {
      cout << "No" << '\n';
      return 0;
    }
  
    if (!val[a]) add(a, b);
    else if (!val[b]) add(b, a);
    else {
      if (a == p[i+1].first || a == p[i+1].second) add(a, b);
      else add(b, a);
    }
  }
  
  cout << "Yes\n";
  for (auto i : ans) cout << i << '\n';
}