#include <iostream>
#include <set>
#include <string>
#include <vector>
using namespace std;
using ll = long long;
int n; ll a, b, c;
vector<string> s;
vector<string> ans;
bool found = false;
void dfs(int id) {
  if (id == n) {
    found = true;
    return;
  }
  if (s[id] == "AB") {
    a++; b--; ans[id] = "A"; if (b >= 0)dfs(id + 1);
    if (found)return; a--; b++;
    b++; a--; ans[id] = "B"; if (a >= 0)dfs(id + 1);
    if (found)return; b--; a++;
  }
  else if (s[id] == "BC") {
    b++; c--; ans[id] = "B"; if (c >= 0)dfs(id + 1);
    if (found)return; b--; c++;
    c++; b--; ans[id] = "C"; if (b >= 0)dfs(id + 1);
    if (found)return; c--; b++;
  }
  else if (s[id] == "AC") {
    c++; a--; ans[id] = "C"; if (a >= 0)dfs(id + 1);
    if (found)return; c--; a++;
    a++; c--; ans[id] = "A"; if (c >= 0)dfs(id + 1);
    if (found)return; a--; c++;
  }
}

int main() {
  cin >> n >> a >> b >> c;
  s.resize(n); ans.resize(n);
  for (int i = 0; i < n; i++)
    cin >> s[i];
  dfs(0);
  if (found) {
    cout << "Yes" << endl;
    for (string ch : ans)cout << ch << endl;
  } else cout << "No" << endl;
  return 0;
}
### 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 <iostream>
#include <set>
#include <string>
#include <vector>
using namespace std;
using ll = long long;
int n; ll a, b, c;
vector<string> s;
vector<string> ans;
bool found = false;
void dfs(int id) {
  if (id == n) {
    found = true;
    return;
  }
  if (s[id] == "AB") {
    a++; b--; ans[id] = "A"; if (b >= 0)dfs(id + 1);
    if (found)return; a--; b++;
    b++; a--; ans[id] = "B"; if (a >= 0)dfs(id + 1);
    if (found)return; b--; a++;
  }
  else if (s[id] == "BC") {
    b++; c--; ans[id] = "B"; if (c >= 0)dfs(id + 1);
    if (found)return; b--; c++;
    c++; b--; ans[id] = "C"; if (b >= 0)dfs(id + 1);
    if (found)return; c--; b++;
  }
  else if (s[id] == "AC") {
    c++; a--; ans[id] = "C"; if (a >= 0)dfs(id + 1);
    if (found)return; c--; a++;
    a++; c--; ans[id] = "A"; if (c >= 0)dfs(id + 1);
    if (found)return; a--; c++;
  }
}

int main() {
  cin >> n >> a >> b >> c;
  s.resize(n); ans.resize(n);
  for (int i = 0; i < n; i++)
    cin >> s[i];
  dfs(0);
  if (found) {
    cout << "Yes" << endl;
    for (string ch : ans)cout << ch << endl;
  } else cout << "No" << endl;
  return 0;
}
```