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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int N = 2e5 + 5;

int n;
string str[N];
char res[N];

void dfs(int i, int a, int b, int c) {
    if (i >= n) {
        cout << "Yes" << endl;
        rep(i,n) {
            cout << res[i] << endl;
        }
        exit(0);
    }
    if (str[i] == "AB") {
        if (a) res[i] = 'B', dfs(i + 1, a - 1, b + 1, c);
        if (b) res[i] = 'A', dfs(i + 1, a + 1, b - 1, c);
    } else if (str[i] == "AC") {
        if (a) res[i] = 'C', dfs(i + 1, a - 1, b, c + 1);
        if (c) res[i] = 'A', dfs(i + 1, a + 1, b, c - 1);
    } else {
        if (b) res[i] = 'C', dfs(i + 1, a, b - 1, c + 1);
        if (c) res[i] = 'B', dfs(i + 1, a, b + 1, c - 1);
    }
}

int main() {
    int a, b, c;
    cin >> n >> a >> b >> c;
    rep(i,n) cin >> str[i];
    dfs(0, a, b, c);
    cout << "No" << endl;
    exit(0);
}