#include <bits/stdc++.h>


typedef long long ll;
using namespace std;


int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,A[3]; 
	cin>>n>>A[0]>>A[1]>>A[2];
    vector<pair<ll,ll>> Q(n);
    for( int i=0;i<n;i++ )
	{
        string s; cin>>s;
        Q[i] = make_pair(s[0]-'A', s[1]-'A');
    }
    vector<char> ans(n);
    for( int i=0;i<n;i++ )
    {
        int x = Q[i].first;
        int y = Q[i].second;
        if ( A[x]==0 && A[y]==0 )
		{
            cout<<"No\n";
            return 0;
        }
        int decide = x;
        if ( A[x]==1 && A[y]==1 )
		{
            if ( i+1<n && ( Q[i+1].first==y || Q[i+1].second==y ) )
                decide = y;
        }
		else if( A[x]>=A[y] )
		{
            decide = y;
        }
        ans[i] = "ABC"[decide];
        
    //    cout<<ans[i]<<endl;
        
		( decide==x ) ? ( A[x]++, A[y]-- ) : ( A[x]--, A[y]++ );
    }
    cout<<"Yes\n";
    for( int i=0;i<n;i++ ) cout<<ans[i]<<"\n";
    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>


typedef long long ll;
using namespace std;


int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,A[3]; 
	cin>>n>>A[0]>>A[1]>>A[2];
    vector<pair<ll,ll>> Q(n);
    for( int i=0;i<n;i++ )
	{
        string s; cin>>s;
        Q[i] = make_pair(s[0]-'A', s[1]-'A');
    }
    vector<char> ans(n);
    for( int i=0;i<n;i++ )
    {
        int x = Q[i].first;
        int y = Q[i].second;
        if ( A[x]==0 && A[y]==0 )
		{
            cout<<"No\n";
            return 0;
        }
        int decide = x;
        if ( A[x]==1 && A[y]==1 )
		{
            if ( i+1<n && ( Q[i+1].first==y || Q[i+1].second==y ) )
                decide = y;
        }
		else if( A[x]>=A[y] )
		{
            decide = y;
        }
        ans[i] = "ABC"[decide];
        
    //    cout<<ans[i]<<endl;
        
		( decide==x ) ? ( A[x]++, A[y]-- ) : ( A[x]--, A[y]++ );
    }
    cout<<"Yes\n";
    for( int i=0;i<n;i++ ) cout<<ans[i]<<"\n";
    return 0;
}

```