#include<bits/stdc++.h>
using namespace std;

#define endl '\n'

int const N = 100010;
string s[N];
int n;
char ans[N];

bool solve(long long i, long long a, long long b, long long c){
    if (a < 0 || b < 0 || c < 0) return false;
    if (i == n) return true;
    if (s[i] == "AB"){
        ans[i] = 'A';
        if (solve(i+1, a+1, b-1, c)) return 1;
        ans[i] = 'B';
        return solve(i+1, a-1, b+1, c);
    } else if (s[i] == "AC"){
        ans[i] = 'A';
        if (solve(i+1, a+1, b, c-1)) return 1;
        ans[i] = 'C';
        return solve(i+1, a-1, b, c+1);
    }
    ans[i] = 'B';
    if (solve(i+1, a, b+1, c-1)) return 1;
    ans[i] = 'C';
    return solve(i+1, a, b-1, c+1);
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    long long a,b,c; cin >> n >> a >> b >> c;
    for (int i=0; i<n; i++) cin >> s[i];
    if (solve(0, a,b,c)){
        cout << "Yes\n";
        for (int i=0; i<n; i++) cout << ans[i] << endl;
    } else cout << "No\n";
}
### Prompt

Please create a solution in Cpp 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;

#define endl '\n'

int const N = 100010;
string s[N];
int n;
char ans[N];

bool solve(long long i, long long a, long long b, long long c){
    if (a < 0 || b < 0 || c < 0) return false;
    if (i == n) return true;
    if (s[i] == "AB"){
        ans[i] = 'A';
        if (solve(i+1, a+1, b-1, c)) return 1;
        ans[i] = 'B';
        return solve(i+1, a-1, b+1, c);
    } else if (s[i] == "AC"){
        ans[i] = 'A';
        if (solve(i+1, a+1, b, c-1)) return 1;
        ans[i] = 'C';
        return solve(i+1, a-1, b, c+1);
    }
    ans[i] = 'B';
    if (solve(i+1, a, b+1, c-1)) return 1;
    ans[i] = 'C';
    return solve(i+1, a, b-1, c+1);
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    long long a,b,c; cin >> n >> a >> b >> c;
    for (int i=0; i<n; i++) cin >> s[i];
    if (solve(0, a,b,c)){
        cout << "Yes\n";
        for (int i=0; i<n; i++) cout << ans[i] << endl;
    } else cout << "No\n";
}
```