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 <iostream>
#include <vector>
#include <algorithm>
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define REP(i, n) for(int i = 1; i < (int)(n); ++i)
using namespace std;

int N;
vector<int> A(3);
vector<char> S(200000), ans(100000);
bool searched = false;

void dfs(int n) {
    if (A[0] < 0 || A[1] < 0 || A[2] < 0) return;
    if (searched) return;
    if (n == N) {
        cout << "Yes" << "\n";
        rep(i, N) cout << ans[i] << "\n";
        searched = true;
        return;
    }
    
    ++A[S[2 * n] - 'A'];
    --A[S[2 * n + 1] - 'A'];
    ans[n] = S[2 * n];
    dfs(n + 1);
    --A[S[2 * n] - 'A'];
    ++A[S[2 * n + 1] - 'A'];
    
    if (searched) return;
    --A[S[2 * n] - 'A'];
    ++A[S[2 * n + 1] - 'A'];
    ans[n] = S[2 * n + 1];
    dfs(n + 1);
    ++A[S[2 * n] - 'A'];
    --A[S[2 * n + 1] - 'A'];
    return;
}

int main() {
    cin >> N >> A[0] >> A[1] >> A[2];
    rep(i, 2 * N) cin >> S[i];
    dfs(0);
    if (!searched) cout << "No" << "\n";
    return 0;
}