#include <bits/stdc++.h>

#define ll long long
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define vll vector<ll>
#define vi vector<int>
#define pi pair<int,int>

using namespace std;

const string str = "ABC";

pi parse(string query)
{
  pi ret;
  if (query=="AB") ret=pi(0,1);
  if (query=="AC") ret=pi(0,2);
  if (query=="BC") ret=pi(1,2);
  return ret;
}

int main()
{
  int N; cin>>N;
  vll v(3);
  rep(i,3) cin>>v[i];
  vector<string> s(N);
  rep(i,N) cin>>s[i];
  if ((v[0]==0&&v[1]==0&&v[2]==0)){
    cout<<"No"<<endl;
    return 0;
  }
  vi ans(N);
  rep(i,N){
    int x,y;
   	tie(x,y) = parse(s[i]);
    if (v[x]==0&&v[y]==0){
      cout<<"No"<<endl;
      return 0;
    }
    if (i<N-1 && v[x]==1&&v[y]==1){
      int nx,ny;
      tie(nx,ny) = parse(s[i+1]);
      if (nx==x||ny==x){
        ans[i] = x;
        v[x]++; v[y]--;
      }else{
        ans[i] = y;
        v[x]--; v[y]++;
      }
    }else{
      if (v[x]>v[y]){
        ans[i] = y;
        v[x]--; v[y]++;
      }else{
        ans[i] = x;
        v[x]++; v[y]--;
      }
    }
  }
  cout<<"Yes"<<endl;
  rep(i,N) cout<<str[ans[i]]<<endl;
  return 0;
}
### Prompt

Please create a solution in CPP to 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 ll long long
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define vll vector<ll>
#define vi vector<int>
#define pi pair<int,int>

using namespace std;

const string str = "ABC";

pi parse(string query)
{
  pi ret;
  if (query=="AB") ret=pi(0,1);
  if (query=="AC") ret=pi(0,2);
  if (query=="BC") ret=pi(1,2);
  return ret;
}

int main()
{
  int N; cin>>N;
  vll v(3);
  rep(i,3) cin>>v[i];
  vector<string> s(N);
  rep(i,N) cin>>s[i];
  if ((v[0]==0&&v[1]==0&&v[2]==0)){
    cout<<"No"<<endl;
    return 0;
  }
  vi ans(N);
  rep(i,N){
    int x,y;
   	tie(x,y) = parse(s[i]);
    if (v[x]==0&&v[y]==0){
      cout<<"No"<<endl;
      return 0;
    }
    if (i<N-1 && v[x]==1&&v[y]==1){
      int nx,ny;
      tie(nx,ny) = parse(s[i+1]);
      if (nx==x||ny==x){
        ans[i] = x;
        v[x]++; v[y]--;
      }else{
        ans[i] = y;
        v[x]--; v[y]++;
      }
    }else{
      if (v[x]>v[y]){
        ans[i] = y;
        v[x]--; v[y]++;
      }else{
        ans[i] = x;
        v[x]++; v[y]--;
      }
    }
  }
  cout<<"Yes"<<endl;
  rep(i,N) cout<<str[ans[i]]<<endl;
  return 0;
}
```