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

Solution (in CPP):

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

vector <string> s;
vector <string> ans; 
int n;
void dfs(int com,int a,int b,int c)
{
    if(a<0 or b<0 or c<0)
        return;
    if(com==n+1){
        cout<<"Yes\n";
        for(int i=1;i<=n;i++)
        {
            cout<<ans[i]<<'\n';
        }
        exit(0);
    }
    if(s[com]=="AB")
    {
        ans[com]="A";
        dfs(com+1,a+1,b-1,c);
        ans[com]="B";
        dfs(com+1,a-1,b+1,c);
    }
    else if(s[com]=="AC")
    {
        ans[com]="A";
        dfs(com+1,a+1,b,c-1);
        ans[com]="C";
        dfs(com+1,a-1,b,c+1);
    }
    else
    {
        ans[com]="B";
        dfs(com+1,a,b+1,c-1);
        ans[com]="C";
        dfs(com+1,a,b-1,c+1);
    }
}


int main()
{
    int i,a,b,c;
    cin>>n>>a>>b>>c;
    s.resize(n+1);
    ans.resize(n+1);
    for(i=1;i<=n;i++)
    {
        cin>>s[i];
    }
    dfs(1,a,b,c);
    cout<<"No";
    return 0;
}