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;
using lint = long long int;
using P = pair<lint, lint>;

int main() {
  lint N, v[3];
  cin >> N >> v[0] >> v[1] >> v[2];

  vector<P> query(N);
  for(lint i=0; i<N; i++) {
    string s;
    cin >> s;

    if(s == "AB") query[i] = {0,1};
    if(s == "AC") query[i] = {0,2};
    if(s == "BC") query[i] = {1,2};
  }
  query.emplace_back(0,1);

  vector<char> ans;
  auto add = [&](lint from, lint to) {
    ans.push_back('A' + to);
    v[to]++;
    v[from]--;
  };

  for(lint i=0; i<N; i++) {
    lint x = query[i].first, y = query[i].second;
    lint X = v[x], Y = v[y];
    if(!X && !Y) {
      cout << "No" << endl;
      return 0;
    } else if(!X) {
      add(y,x);
    } else if(!Y){
      add(x,y);
    } else {
      if(x == query[i+1].first || x == query[i+1].second) {
        add(y,x);
      } else {
        add(x,y);
      }
    }
  }

  cout << "Yes" << endl;
  for(auto e : ans) cout << e << endl;
}