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>
using namespace std;

int main() {
    long n, a, b, c;
    cin >> n >> a >> b >> c;
    string t, u;
    for(int i = 0; i < n; i++) {
        string s;
        cin >> s;
        t += s == "AB" ? 'C' : s == "AC" ? 'B' : 'A';
    }
    for(int i = 0; i < n; i++) {
        switch(t[i]) {
        case 'A':
            if(!c || b && i + 1 < n && t[i + 1] == 'B')
                b--, c++, u += 'C';
            else
                b++, c--, u += 'B';
            break;
        case 'B':
            if(!c || a && i + 1 < n && t[i + 1] == 'A')
                a--, c++, u += 'C';
            else
                a++, c--, u += 'A';
            break;
        default:
            if(!b || a && i + 1 < n && t[i + 1] == 'A')
                a--, b++, u += 'B';
            else
                a++, b--, u += 'A';
        }
        if(a < 0 || b < 0 || c < 0) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes\n";
    for(auto y : u)
        cout << y << '\n';
    return 0;
}