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;

int main()
{
  int n;
  vector<int64_t> m(3);
  cin >> n >> m[0] >> m[1] >> m[2];
  vector<pair<int, int>> vp;
  for (int i = 0; i < n; i++)
  {
    char a, b;
    cin >> a >> b;
    vp.emplace_back((int)(a - 'A'), (int)(b - 'A'));
  }
  vp.emplace_back(0, 1);
  vector<int> ans;
  auto add = [&](int a, int b) {
    m[a]++, m[b]--;
    ans.push_back(a);
  };
  for (int i = 0; i < n; i++)
  {
    int a, b, na, nb;
    tie(a, b) = vp[i];
    if (!m[a] && !m[b])
    {
      cout << "No" << endl;
      return 0;
    }
    else if (!m[a])
    {
      add(a, b);
    }
    else if (!m[b])
    {
      add(b, a);
    }
    else
    {
      tie(na, nb) = vp[i + 1];
      if (a == na || a == nb)
      {
        add(a, b);
      }
      else
      {
        add(b, a);
      }
    }
  }
  cout << "Yes" << endl;
  for (int i = 0; i < n; i++)
  {
    cout << (char)(ans[i] + 'A') << endl;
  }
}