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 <vector>
#include <iostream>
using namespace std;
int main() {
    int n, a[3]; cin >> n >> a[0] >> a[1] >> a[2];
    string s[n]; for (int i = 0; i < n; i++) cin >> s[i];
    vector<char> ans;
    for (int i = 0; i < n; i++) {
        int p = s[i][0]-'A', q = s[i][1]-'A';
        if (a[p] == 0 && a[q] == 0) {
            cout << "No\n";
            return 0;
        } else if (a[p] == 0) {
            ans.push_back(s[i][0]);
            a[p]++;
            a[q]--;
        } else if (a[q] == 0) {
            ans.push_back(s[i][1]);
            a[p]--;
            a[q]++;
        } else {
            if (i == n-1) ans.push_back(s[i][1]);
            else {
                int p1 = s[i+1][0] - 'A', q1 = s[i+1][1] - 'A';
                for (int j = 0; j < 3; j++)
                    if ((p == j || q == j) && (p1 == j || q1 == j)) {
                        ans.push_back('A' + j);
                        a[j]++;
                        a[p+q-j]--;
                        break;
                    }
            }
        }
    }
    cout << "Yes\n";
    for (char c : ans) cout << c << endl;
}