#include <iostream>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, a[3];
    cin >> n >> a[0] >> a[1] >> a[2];
    int x1[n], x2[n];
    for (int i = 0; i < n; i++) {
        char c1, c2;
        cin >> c1 >> c2;
        x1[i] = c1 - 'A';
        x2[i] = c2 - 'A';
    }

    int ans[n];
    for (int i = 0; i < n; i++) {
        if (a[x1[i]] == 0 && a[x2[i]] == 0) {
            cout << "No" << endl;
            return 0;
        }

        if (a[x1[i]] == 1 && a[x2[i]] == 1 && a[3-x1[i]-x2[i]] == 0 && i < n - 1) {
            if (x1[i] != x1[i+1] || x2[i] != x2[i+1]) {
                ans[i] = x1[i] == x1[i+1] || x1[i] == x2[i+1] ? x1[i] : x2[i];
            } else {
                ans[i] = x1[i];
            }
        } else {
            ans[i] = a[x1[i]] < a[x2[i]] ? x1[i] : x2[i];
        }

        a[ans[i]]++;
        a[x1[i]+x2[i]-ans[i]]--;
    }

    cout << "Yes" << endl;
    for (int i = 0; i < n; i++) {
        cout << (char)(ans[i] + 'A') << endl;
    }

    return 0;
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <iostream>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, a[3];
    cin >> n >> a[0] >> a[1] >> a[2];
    int x1[n], x2[n];
    for (int i = 0; i < n; i++) {
        char c1, c2;
        cin >> c1 >> c2;
        x1[i] = c1 - 'A';
        x2[i] = c2 - 'A';
    }

    int ans[n];
    for (int i = 0; i < n; i++) {
        if (a[x1[i]] == 0 && a[x2[i]] == 0) {
            cout << "No" << endl;
            return 0;
        }

        if (a[x1[i]] == 1 && a[x2[i]] == 1 && a[3-x1[i]-x2[i]] == 0 && i < n - 1) {
            if (x1[i] != x1[i+1] || x2[i] != x2[i+1]) {
                ans[i] = x1[i] == x1[i+1] || x1[i] == x2[i+1] ? x1[i] : x2[i];
            } else {
                ans[i] = x1[i];
            }
        } else {
            ans[i] = a[x1[i]] < a[x2[i]] ? x1[i] : x2[i];
        }

        a[ans[i]]++;
        a[x1[i]+x2[i]-ans[i]]--;
    }

    cout << "Yes" << endl;
    for (int i = 0; i < n; i++) {
        cout << (char)(ans[i] + 'A') << endl;
    }

    return 0;
}

```