#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n,a,b,c; cin>>n>>a>>b>>c;
    int v[3]={a,b,c};
    string ch="ABC";
    vector<pair<int,int>> ope(n);
    for(int i=0; i<n; ++i)
    {
        string s; cin>>s;
        ope[i]={s[0]-'A',s[1]-'A'};
    }
    string ans;
    stack<int> bac;

    for(auto i=ope.begin(); i!=ope.end(); ++i)
    {
        auto [x,y]=*i;

        if(v[x]>v[y])
        {
            ans+=ch[y];
            v[x]--;
            v[y]++;
        }
        else
        {
            if(v[x]==v[y] and v[x]==1 and next(i)!=end(ope))
            {
                auto ii=next(i);
                if(y==ii->first or y==ii->second)
                {
                    swap(x,y);
                }
            }
            if(!v[y])
            {
                cout << "No" << "\n";
                return 0;
            }
            v[x]++;
            v[y]--;
            ans+=ch[x];
        }
    }

    cout << "Yes" << "\n";
    for(char c: ans) cout << c << "\n";
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
using namespace std;

int main()
{
    int n,a,b,c; cin>>n>>a>>b>>c;
    int v[3]={a,b,c};
    string ch="ABC";
    vector<pair<int,int>> ope(n);
    for(int i=0; i<n; ++i)
    {
        string s; cin>>s;
        ope[i]={s[0]-'A',s[1]-'A'};
    }
    string ans;
    stack<int> bac;

    for(auto i=ope.begin(); i!=ope.end(); ++i)
    {
        auto [x,y]=*i;

        if(v[x]>v[y])
        {
            ans+=ch[y];
            v[x]--;
            v[y]++;
        }
        else
        {
            if(v[x]==v[y] and v[x]==1 and next(i)!=end(ope))
            {
                auto ii=next(i);
                if(y==ii->first or y==ii->second)
                {
                    swap(x,y);
                }
            }
            if(!v[y])
            {
                cout << "No" << "\n";
                return 0;
            }
            v[x]++;
            v[y]--;
            ans+=ch[x];
        }
    }

    cout << "Yes" << "\n";
    for(char c: ans) cout << c << "\n";
}
```