#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<int> d(3);
  rep(i,3) cin >> d[i];
  vector<P> q;
  rep(i,n) {
    string s;
    cin >> s;
    int a = 0, b = 1;
    if (s == "AB") a = 0, b = 1;
    if (s == "AC") a = 0, b = 2;
    if (s == "BC") a = 1, b = 2;
    q.emplace_back(a,b);
  }
  string ans;
  auto add = [&](int a, int b) {
    d[a]++;
    d[b]--;
    ans += 'A'+a;
  };
  q.emplace_back(0,1);
  rep(i,n) {
    int a = q[i].first;
    int b = q[i].second;
    if (!d[a] && !d[b]) {
      cout << "No" << endl;
      return 0;
    }
    if (!d[a]) {
      add(a,b);
    } else if (!d[b]) {
      add(b,a);
    } else {
      int na = q[i+1].first;
      int nb = q[i+1].second;
      if (a == na || a == nb) {
        add(a,b);
      } else {
        add(b,a);
      }
    }
  }
  cout << "Yes" << endl;
  rep(i,n) {
    cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<int> d(3);
  rep(i,3) cin >> d[i];
  vector<P> q;
  rep(i,n) {
    string s;
    cin >> s;
    int a = 0, b = 1;
    if (s == "AB") a = 0, b = 1;
    if (s == "AC") a = 0, b = 2;
    if (s == "BC") a = 1, b = 2;
    q.emplace_back(a,b);
  }
  string ans;
  auto add = [&](int a, int b) {
    d[a]++;
    d[b]--;
    ans += 'A'+a;
  };
  q.emplace_back(0,1);
  rep(i,n) {
    int a = q[i].first;
    int b = q[i].second;
    if (!d[a] && !d[b]) {
      cout << "No" << endl;
      return 0;
    }
    if (!d[a]) {
      add(a,b);
    } else if (!d[b]) {
      add(b,a);
    } else {
      int na = q[i+1].first;
      int nb = q[i+1].second;
      if (a == na || a == nb) {
        add(a,b);
      } else {
        add(b,a);
      }
    }
  }
  cout << "Yes" << endl;
  rep(i,n) {
    cout << ans[i] << endl;
  }
  return 0;
}

```