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>
typedef long long ll;
using namespace std;
int a[3];
bool contains(string s, char c){
  return s[0] == c || s[1] == c;
}
int main()
{
  int n;
  cin >> n >> a[0] >> a[1] >> a[2];
  vector<string> qs(n);
  for(auto& el : qs)
    cin >> el;
//  return 0;

  vector<int> nxt(n);
  nxt[n - 1] = n;
  for(int i = n - 2;i >= 0;i--){
    if(qs[i] == qs[i + 1])
      nxt[i] = nxt[i + 1];
    else
      nxt[i] = i + 1;
  }
  vector<char> ans;
  for(int i = 0;i < n;i++){
    string q = qs[i];
    char x = q[0], y = q[1];
    int &f = a[x - 'A'], &s = a[y - 'A'];
    if(f == s && f == 0){
      cout << "No";
      return 0;
    }
    if(f < s || (f == s && (nxt[i] == n || contains(qs[nxt[i]], x)))){
      f++, s--;
      ans.push_back(x);
    }else{
      f--, s++;
      ans.push_back(y);
    }
  }
  cout << "Yes\n";
  for(auto el : ans)
    cout << el << endl;
  return 0;
}