typedef long long ll;
typedef long double ld;
#include <bits/stdc++.h>
using namespace std;

int main() {
    ll n;
    std::cin >> n;
    map<char,ll> val;
    for (int i = 'A'; i <= 'C'; i++) {
        std::cin >> val[i];
    }
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
        std::cin >> s[i];
    }
    vector<char> ans;
    for (int i = 0; i < n; i++) {
        char c1 = s[i][0];
        char c2 = s[i][1];
        char cL = ((val[c1]>=val[c2]) ? c1 : c2 );
        char cS = c1^c2^cL;
        if(val[cL]==0){
            std::cout << "No" << std::endl;
            return 0;
        }
        if(val[cL]==1&&val[cS]==1&&i+1<n&&(s[i+1][0]==cL||s[i+1][1]==cL))swap(cL,cS);
        ans.push_back(cS);
        val[cL]--;
        val[cS]++;
        // std::cout << val['A']<<" "<<val['B']<<" "<<val['C'] << std::endl;
    }
    std::cout << "Yes" << std::endl;
    for (auto e : ans) {
        std::cout << e << std::endl;
    }
}
### Prompt

Create a solution in cpp for 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
typedef long long ll;
typedef long double ld;
#include <bits/stdc++.h>
using namespace std;

int main() {
    ll n;
    std::cin >> n;
    map<char,ll> val;
    for (int i = 'A'; i <= 'C'; i++) {
        std::cin >> val[i];
    }
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
        std::cin >> s[i];
    }
    vector<char> ans;
    for (int i = 0; i < n; i++) {
        char c1 = s[i][0];
        char c2 = s[i][1];
        char cL = ((val[c1]>=val[c2]) ? c1 : c2 );
        char cS = c1^c2^cL;
        if(val[cL]==0){
            std::cout << "No" << std::endl;
            return 0;
        }
        if(val[cL]==1&&val[cS]==1&&i+1<n&&(s[i+1][0]==cL||s[i+1][1]==cL))swap(cL,cS);
        ans.push_back(cS);
        val[cL]--;
        val[cS]++;
        // std::cout << val['A']<<" "<<val['B']<<" "<<val['C'] << std::endl;
    }
    std::cout << "Yes" << std::endl;
    for (auto e : ans) {
        std::cout << e << std::endl;
    }
}
```