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;

const int maxn = 1e5 + 5;
char s[maxn][3];
int a[3], n;

signed main() {
  // freopen("in", "r", stdin);
  scanf("%d%d%d%d", &n, &a[0], &a[1], &a[2]);
  for(int i = 0; i < n; i++) {
    scanf("%s", s[i]);
  }
  vector<char> ans;
  bool flag = true;
  for(int i = 0; i < n; i++) {
    int ta = s[i][0] - 'A', tb = s[i][1] - 'A';
    if(a[ta] > a[tb]) {
      ans.push_back(s[i][1]);
      a[tb]++, a[ta]--;
    }else if(a[tb] > a[ta]){
      ans.push_back(s[i][0]);
      a[ta]++, a[tb]--;
    }else {
      if(a[ta] == 0) {
        flag = false;
        break;
      }
      if(i + 1 >= n || s[i][0] == s[i + 1][0] || s[i][0] == s[i + 1][1]) {
        ans.push_back(s[i][0]);
        a[ta]++, a[tb]--;
      }else {
        ans.push_back(s[i][1]);
        a[tb]++, a[ta]--;
      }
    }
  }
  if(flag) {
    printf("Yes\n");
    for(int i = 0; i < n; i++) {
      printf("%c\n", ans[i]);
    }
  }else {
    printf("No\n");
  }
  return 0;
}