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 fail(){
    cout << "No" << endl;
    exit(0);
}

int main(){
    int N, A[3];
    cin >> N >> A[0] >> A[1] >> A[2];
    vector<vector<int>> S(N);
    for(int i=0; i<N; i++){
        string s;
        cin >> s;
        for(char c : s) S[i].push_back(c-'A');
    }

    vector<vector<int>> nxt(N, vector<int>(3, N));
    vector<int> now(3, N);
    for(int i=N-1; i>=0; i--){
        nxt[i] = now;
        for(int j : S[i]) now[j] = i;
    }

    vector<int> ans;
    for(int i=0; i<N; i++){
        int a = S[i][0], b = S[i][1];
        if(A[a] == 0 && A[b] == 0) fail();
        if(A[a] > A[b] || (A[a] == A[b] && nxt[i][a] > nxt[i][b])) swap(a, b);
        ans.push_back(a);
        A[a]++;
        A[b]--;
    }

    cout << "Yes" << endl;
    for(int a : ans) cout << char('A'+a) << endl;
    return 0;
}