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

int main() {
  int n;
  cin >> n;
  map<char, int> cnt;
  rep (i, 3) cin >> cnt['A'+i];
  vector<string> s(n);
  for (string &i : s) cin >> i;
  string ans;
  rep (i, n) {
    char t1 = s[i][0], t2 = s[i][1];
    if (cnt[t1] == 0 and cnt[t2] == 0) {
      cout << "No" << endl;
      return 0;
    }
    if (cnt[t1] < cnt[t2]) swap(t1, t2);
    char t3 = 'A'+3-(t1-'A')-(t2-'A');
    //while (t3 == t1 or t3 == t2) t3++;
    if (cnt[t1] == 1 and cnt[t2] == 1 and cnt[t3] == 0) {
      if (i + 1 < n) {
        if (t1 == s[i+1][0] or t1 == s[i+1][1]) {
          swap(t1, t2);
        }
      }
    }
    ans += t2;
    cnt[t2]++;
    cnt[t1]--;
  }

  cout << "Yes" << endl;
  for (char i : ans) cout << i << endl;
  return 0;
}