#include <bits/stdc++.h>
using namespace std;
#define int long
void fail() {
    cout << "No";
    exit(0);
}
signed main() {
    int n;
    map<char, int> var;
    cin >> n >> var['A'] >> var['B'] >> var['C'];
    vector<string> str(n);
    vector<char> res(n);
    for (auto& i : str) cin >> i;
    for (int i = 0; i < n; ++i) {
        char c0 = str[i][0], c1 = str[i][1];
        if (!var[c0] && !var[c1]) fail();
        if (!var[c0])
            ++var[c0], --var[c1], res[i] = c0;
        else if (!var[c1])
            ++var[c1], --var[c0], res[i] = c1;
        else {
            if (i + 1 < n && str[i + 1] != str[i]) {
                if (str[i + 1][0] == c0 || str[i + 1][1] == c0)
                    ++var[c0], --var[c1], res[i] = c0;
                else
                    ++var[c1], --var[c0], res[i] = c1;
            } else {
                if (var[c0] < var[c1])
                    ++var[c0], --var[c1], res[i] = c0;
                else
                    ++var[c1], --var[c0], res[i] = c1;
            }
        }
    }
    cout << "Yes" << endl;
    for (auto ch : res) cout << ch << endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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;
#define int long
void fail() {
    cout << "No";
    exit(0);
}
signed main() {
    int n;
    map<char, int> var;
    cin >> n >> var['A'] >> var['B'] >> var['C'];
    vector<string> str(n);
    vector<char> res(n);
    for (auto& i : str) cin >> i;
    for (int i = 0; i < n; ++i) {
        char c0 = str[i][0], c1 = str[i][1];
        if (!var[c0] && !var[c1]) fail();
        if (!var[c0])
            ++var[c0], --var[c1], res[i] = c0;
        else if (!var[c1])
            ++var[c1], --var[c0], res[i] = c1;
        else {
            if (i + 1 < n && str[i + 1] != str[i]) {
                if (str[i + 1][0] == c0 || str[i + 1][1] == c0)
                    ++var[c0], --var[c1], res[i] = c0;
                else
                    ++var[c1], --var[c0], res[i] = c1;
            } else {
                if (var[c0] < var[c1])
                    ++var[c0], --var[c1], res[i] = c0;
                else
                    ++var[c1], --var[c0], res[i] = c1;
            }
        }
    }
    cout << "Yes" << endl;
    for (auto ch : res) cout << ch << endl;
}
```