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;
using ll = long long;
using P = pair<int,int>;
#define rep(i, n) for(int i=0; i<(int)(n); ++i)

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  vector<int> d(3);
  vector<P> q;
  cin >> n;
  rep(i, 3) cin >> d[i];
  rep(i, n) {
    string s;
    cin >> s;
    q.emplace_back(s[0]-'A', s[1]-'A');
  }
  q.emplace_back(0, 0);
  string ans;
  auto add = [&] (int a, int b) {
    d[a]++;
    d[b]--;
    ans.push_back('A'+a);
  };
  rep(i, n) {
    int q1 = q[i].first;
    int q2 = q[i].second;
    if (!d[q1] && !d[q2]) {
      cout << "No\n";
      return 0;
    }
    if (!d[q1]) add(q1, q2);
    else if (!d[q2]) add(q2, q1);
    else {
      int nx1 = q[i+1].first;
      int nx2 = q[i+1].second;
      if (q1 == nx1 || q1 == nx2) add(q1, q2);
      else add(q2, q1);
    }
  }
  cout << "Yes\n";
  rep(i, ans.size()) cout << ans[i] << endl;
}