#include <bits/stdc++.h>
#define rep(i, z, n) for(ll i = z; i <= n; i++)
#define repl(i, z, n) for(ll i = z; i < n; i++)
#define all(v) v.begin(), v.end()
typedef long long ll;

using namespace std;

int main(){
  vector<int> a=vector<int>(3);
  vector<int> res;
  int n;
  cin >> n;
  vector<string> s=vector<string>(n);
  repl(i,0,3) cin >> a[i];
  repl(i,0,n) cin >> s[i];
  
  if(accumulate(all(a),0)==0){
    cout << "No\n";
    exit(0);
  }
  repl(i,0,n){
    int x = s[i][0]-'A';
    int y = s[i][1]-'A';
    if(i<n-1 && a[x]==1 && a[y]==1){
      bool found_x=(x==(s[i+1][0]-'A') || x==(s[i+1][1]-'A'));
      if(found_x){
        a[x]++;
        a[y]--;
        res.push_back(x); 
      }else{
        a[x]--;
        a[y]++;
        res.push_back(y); 
      }
    }else{
      if(a[x]<=0 && a[y] <=0){
        cout << "No\n";
        exit(0);
      }
      if(a[x]<a[y]){
        a[x]++;
        a[y]--;
        res.push_back(x);
      }else{
        a[y]++;
        a[x]--;
        res.push_back(y);

      }
    } 
  }
  cout << "Yes\n";
  for(int i:res){
    cout << char(i+'A') << 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, z, n) for(ll i = z; i <= n; i++)
#define repl(i, z, n) for(ll i = z; i < n; i++)
#define all(v) v.begin(), v.end()
typedef long long ll;

using namespace std;

int main(){
  vector<int> a=vector<int>(3);
  vector<int> res;
  int n;
  cin >> n;
  vector<string> s=vector<string>(n);
  repl(i,0,3) cin >> a[i];
  repl(i,0,n) cin >> s[i];
  
  if(accumulate(all(a),0)==0){
    cout << "No\n";
    exit(0);
  }
  repl(i,0,n){
    int x = s[i][0]-'A';
    int y = s[i][1]-'A';
    if(i<n-1 && a[x]==1 && a[y]==1){
      bool found_x=(x==(s[i+1][0]-'A') || x==(s[i+1][1]-'A'));
      if(found_x){
        a[x]++;
        a[y]--;
        res.push_back(x); 
      }else{
        a[x]--;
        a[y]++;
        res.push_back(y); 
      }
    }else{
      if(a[x]<=0 && a[y] <=0){
        cout << "No\n";
        exit(0);
      }
      if(a[x]<a[y]){
        a[x]++;
        a[y]--;
        res.push_back(x);
      }else{
        a[y]++;
        a[x]--;
        res.push_back(y);

      }
    } 
  }
  cout << "Yes\n";
  for(int i:res){
    cout << char(i+'A') << endl;
  }
  return 0; 
}


```