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;

typedef long long ll;

ll N, V[3];
vector<ll> K;
vector<ll> R;

bool solve() {
  for (ll i = 0; i < N; ++i) {
    ll k = K[i];
    ll a = (k + 1) % 3, b = (k + 2) % 3;
    if (V[a] == 0 && V[b] == 0) return false;
    if (V[a] == 1 && V[b] == 1 && i + 1 < N && K[i + 1] != k) {
      ll c = K[i + 1];
      ll d = c == a ? b : a;
      ++V[d];
      --V[c];
      R.push_back(d);
    } else if (V[a] < V[b]) {
      ++V[a];
      --V[b];
      R.push_back(a);
    } else {
      ++V[b];
      --V[a];
      R.push_back(b);
    }
  }
  return true;
}
int main() {
  cin >> N >> V[0] >> V[1] >> V[2];
  K.resize(N);
  for (ll i = 0; i < N; ++i) {
    string s;
    cin >> s;
    K[i] = s == "AB" ? 2 : s == "AC" ? 1 : 0;
  }
  if (!solve()) {
    cout << "No" << endl;
  } else {
    cout << "Yes" << endl;
    for (ll r : R) {
      cout << (r == 0 ? "A" : r == 1 ? "B" : "C") << endl;
    }
  }
  return 0;
}