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;
#define all(x) (x).begin(),(x).end()
const long long mod = 1e9+7;
//////////////////////////////////

int main() {
  int x[3];
  int n; cin >> n >> x[0] >> x[1] >> x[2];
  vector<string> s(n);
  string ans = "";
  for(int i=0; i<n; i++) cin >> s[i];
  for(int i=0; i<n; i++) {
    int p = s[i][0]-'A';
    int q = s[i][1]-'A';
    if (x[p] == 0 && x[q] == 0) {
      cout << "No" << endl;
      return 0;
    }
    if (x[p] > x[q]) {
      char c = q + 'A'; x[p]--; x[q]++;
      ans += c;
    } else if(x[p] < x[q]) {
      char c = p + 'A'; x[q]--; x[p]++;
      ans += c;
    } else {
      if (x[p] >= 2 || i == n-1) {
        char c = p + 'A'; x[q]--; x[p]++;
        ans += c;
      } else {
        int np = s[i+1][0]-'A';
        int nq = s[i+1][1]-'A';
        if (p == np || p == nq ) {
          char c = p + 'A'; x[q]--; x[p]++;
          ans += c;
        } else {
          char c = q + 'A'; x[p]--; x[q]++;
          ans += c;
        }
      }
    }
  }

  cout << "Yes" << endl;
  for(int i=0; i<n; i++) cout << ans[i] << endl;

  return 0;
}

// EOF