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;
char s[100010][2];
int a[3];
vector<char>ans;
int main(){
    int n; scanf("%d%d%d%d", &n, &a[0], &a[1], &a[2]);
    for (int i = 1; i <= n; i++) scanf(" %s", s[i]), s[i][0] -= 'A', s[i][1] -= 'A';
    for (int i = 1; i <= n; i++){
        char cx = s[i][0], cy = s[i][1];
        int &x = a[cx];
        int &y = a[cy];
        if (x == 0 && y == 0){
            puts("No");
            return 0;
        }
        if (x > y) x--, y++, ans.push_back('A' + cy);
        else if (x < y) y--, x++, ans.push_back('A' + cx);
        else {
            char nx = s[i + 1][0], ny = s[i + 1][1];
            if (i == n || (i < n && nx == cx && ny == cy)) x--, y++, ans.push_back('A' + s[i][1]);
            else {
                if (cx == nx || cx == ny) y--, x++, ans.push_back('A' + cx);
                else if (cy == nx || cy == ny) x--, y++, ans.push_back('A' + cy);
            }
        }
        assert(a[0] >= 0 && a[1] >= 0 && a[2] >= 0);
    }
    puts("Yes");
    for (char ch : ans) printf("%c\n", ch);
    return 0;
}