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>

using namespace std;

#define rep(i,n) for(int i=0; i<(int)(n); i++)

int main() {
  int N, v[3];
  cin >> N >> v[0] >> v[1] >> v[2];
  string s[N+1];
  rep(i,N) cin >> s[i];
  s[N] = "XX";
  
  int b = 1, r[N];
  rep(i,N) {
    int x = s[i][0] - 'A', y = s[i][1] - 'A';
    if(v[x] == 0 && v[y] == 0) b = 0;
    else if(v[x] == 0) { r[i] = x; v[x]++; v[y]--; }
    else if(v[y] == 0) { r[i] = y; v[y]++; v[x]--; }
    else if(v[x] == 1 && v[y] == 1) {
      int j = i+1;
      for(; s[i] == s[j]; j++);
      int nx = s[j][0] - 'A', ny = s[j][1] - 'A';
      if(x == nx || x == ny) { r[i] = x; v[x]++; v[y]--; }
      else { r[i] = y; v[y]++; v[x]--; }
    }
    else if(v[x] > 1) { r[i] = y; v[y]++; v[x]--; }
    else { r[i] = x; v[x]++; v[y]--; }
  }
  
  if(b) {
    cout << "Yes" << endl;
    rep(i,N) cout << (char)(r[i] + 'A') << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}