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;

int main(void) {
  int n, a[3]; cin >> n >> a[0] >> a[1] >> a[2];
  int ans[n], x[n][2];
  
  for(int i = 0;i < n;i++) {
    char s[3]; cin >> s; 
    for(int j = 0;j < 2;j++) { x[i][j] = s[j] - 'A'; } 
  }
  
  for(int i = 0;i < n;i++) {
    if( a[ x[i][0] ] == 0 && a[ x[i][1] ] == 0 ) { cout << "No\n"; return 0; }
    else if( a[ x[i][0] ] == 0 ) { a[ x[i][0] ]++; a[ x[i][1] ]--; ans[i] = x[i][0]; }
    else if( a[ x[i][1] ] == 0 ) { a[ x[i][1] ]++; a[ x[i][0] ]--; ans[i] = x[i][1]; }
    else {
      if( i < n - 1 && ( x[i][0] == x[i + 1][0] || x[i][0] == x[i + 1][1] ) ) {
        a[ x[i][0] ]++; a[ x[i][1] ]--; ans[i] = x[i][0];
      } else { a[ x[i][1] ]++; a[ x[i][0] ]--; ans[i] = x[i][1]; }
    }
  }
  
  cout << "Yes\n"; for(int i = 0;i < n;i++) { cout << (char)( ans[i] + 'A' ) << endl; }
  return 0;
}