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;

long long n, d[3] = {}, sum;
string res;
vector<string> s;

bool solve();

int main() {
  cin >> n;
  for (int i = 0; i < 3; ++i) cin >> d[i];
  s.resize(n);
  for (int i = 0; i < n; ++i) cin >> s[i];
  if (solve()) {
    cout << "Yes" << endl;
    for (int i = 0; i < n; ++i) cout << res[i] << endl;
  } else
    cout << "No" << endl;
  return 0;
}

bool solve() {
  sum = d[0] + d[1] + d[2];
  if (sum == 0) return 0;
  for (int i = 0; i < n; ++i) {
    string now = s[i];
    long long x[2] = {now[0] - 'A', now[1] - 'A'}, p = 0;
    if (d[x[0]] > d[x[1]])
      p = 0;
    else if (d[x[0]] < d[x[1]])
      p = 1;
    else if (d[x[0]] == 0)
      return 0;
    else {
      if (i + 1 < n && s[i] != s[i + 1]) {
        long long y[2] = {s[i + 1][0] - 'A', s[i + 1][1] - 'A'};
        if (x[0] == y[0] || x[0] == y[1]) p = 1;
      }
    }
    --d[x[p]];
    ++d[x[1 - p]];
    res += x[1 - p] + 'A';
  }
  return 1;
}