#include <bits/stdc++.h>

using namespace std;

int main(){
    int64_t n;
    vector<int64_t> A(3);
    cin >> n;
    for (auto &a: A) cin >> a;
    vector<string> S(n);
    copy_n(istream_iterator<string>(cin), n, S.begin());
    vector<array<int, 2>> B(n);
    for (int i = 0; i < n; i++){
        if (S[i] == "AB")
            B[i] = {0, 1};
        if (S[i] == "AC")
            B[i] = {0, 2};
        if (S[i] == "BC")
            B[i] = {1, 2};
    }

    vector<int> ans;

    for (int i = 0; i < n; i++){
        int j = B[i][0], k = B[i][1];
        if (A[j] == 1 and A[k] == 1 and i < n - 1){
            if (!(B[i + 1][0] == j or B[i + 1][1] == j))
                swap(j, k);
        } else {
            if (A[j] > A[k])
                swap(j, k);
        }

        ans.push_back(j);
        A[j]++; A[k]--;

        if (A[j] < 0 or A[k] < 0){
            puts("No");
            return 0;
        }
    }

    puts("Yes");
    for (const auto &a: ans){
        cout << char(a + 'A') << endl;
    }
}
### Prompt

Your task is to create 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 <bits/stdc++.h>

using namespace std;

int main(){
    int64_t n;
    vector<int64_t> A(3);
    cin >> n;
    for (auto &a: A) cin >> a;
    vector<string> S(n);
    copy_n(istream_iterator<string>(cin), n, S.begin());
    vector<array<int, 2>> B(n);
    for (int i = 0; i < n; i++){
        if (S[i] == "AB")
            B[i] = {0, 1};
        if (S[i] == "AC")
            B[i] = {0, 2};
        if (S[i] == "BC")
            B[i] = {1, 2};
    }

    vector<int> ans;

    for (int i = 0; i < n; i++){
        int j = B[i][0], k = B[i][1];
        if (A[j] == 1 and A[k] == 1 and i < n - 1){
            if (!(B[i + 1][0] == j or B[i + 1][1] == j))
                swap(j, k);
        } else {
            if (A[j] > A[k])
                swap(j, k);
        }

        ans.push_back(j);
        A[j]++; A[k]--;

        if (A[j] < 0 or A[k] < 0){
            puts("No");
            return 0;
        }
    }

    puts("Yes");
    for (const auto &a: ans){
        cout << char(a + 'A') << endl;
    }
}
```