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;


void solve() {
    int n; cin >> n;
    vector<int> a(3);
    for (auto& x: a) cin >> x;
    vector<string> s(n);
    for (auto& x: s) cin >> x;
    string res(n,'?');
    for (int _ = 0; _ < n; _++) {
        int i = s[_][0]-'A';
        int j = s[_][1]-'A';
        if (a[i] == 0 && a[i]==a[j]) {
            cout << "No"; return;
        }
        int k = i;
        if (a[i] == 1 && a[i]==a[j]) {
            if (_+1<n) {
                int ii = s[_+1][0]-'A';
                int jj = s[_+1][1]-'A';
                if (i==ii||i==jj) k = i;
                else k = j;
            }
        } else {
            k = a[i]>a[j]? j : i;
        }
        res[_] = char('A'+k);
        ++a[k]; --a[k==i?j:i];
    }
    cout << "Yes\n";
    for (char c:res) cout << c << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}