//author : raghav_0901

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

int n;
int arr[3];
signed main()
{
    cin>>n>>arr[0]>>arr[1]>>arr[2];
    vector<string> s(n);
    for(auto &x:s) cin>>x;
    vector<char> ans;
    for(int i=0;i<n;i++)
    {
        int p1=s[i][0]-'A';
        int p2=s[i][1]-'A';
        if(arr[p1]+arr[p2]==0)
        {
            cout<<"No"<<endl;
            return 0;
        }
        else if(arr[p1]==arr[p2] && i+1<n)
        {
            int np1=s[i+1][0]-'A';
            int np2=s[i+1][1]-'A';
            if(np1==p1 || np2==p1)
            {
                ans.push_back(s[i][0]);
                arr[p1]++;arr[p2]--;
            }
            else
            {
                ans.push_back(s[i][1]);
                arr[p1]--;arr[p2]++;
            }
        }
        else if(arr[p1]<arr[p2])
        {
            ans.push_back(s[i][0]);
            arr[p1]++;arr[p2]--;
        }
        else
        {
            ans.push_back(s[i][1]);
            arr[p1]--;arr[p2]++;
        }
    }
    cout<<"Yes"<<endl;
    for(auto x:ans) cout<<x<<'\n';
}

### Prompt

Your challenge is to write a Cpp solution 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
//author : raghav_0901

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

int n;
int arr[3];
signed main()
{
    cin>>n>>arr[0]>>arr[1]>>arr[2];
    vector<string> s(n);
    for(auto &x:s) cin>>x;
    vector<char> ans;
    for(int i=0;i<n;i++)
    {
        int p1=s[i][0]-'A';
        int p2=s[i][1]-'A';
        if(arr[p1]+arr[p2]==0)
        {
            cout<<"No"<<endl;
            return 0;
        }
        else if(arr[p1]==arr[p2] && i+1<n)
        {
            int np1=s[i+1][0]-'A';
            int np2=s[i+1][1]-'A';
            if(np1==p1 || np2==p1)
            {
                ans.push_back(s[i][0]);
                arr[p1]++;arr[p2]--;
            }
            else
            {
                ans.push_back(s[i][1]);
                arr[p1]--;arr[p2]++;
            }
        }
        else if(arr[p1]<arr[p2])
        {
            ans.push_back(s[i][0]);
            arr[p1]++;arr[p2]--;
        }
        else
        {
            ans.push_back(s[i][1]);
            arr[p1]--;arr[p2]++;
        }
    }
    cout<<"Yes"<<endl;
    for(auto x:ans) cout<<x<<'\n';
}

```