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

int n;
int a[3]; 
string s[100001];
string ans;

bool ok() {
  for (int i = 0; i < n; ++i) {
    int p = s[i][0] - 'A';
    int q = s[i][1] - 'A';
    if (a[p] == 0 && a[q] == 0) return false;
    int ch;
    if (a[p] == 0) {
      ch = p;
    } else if (a[q] == 0) {
      ch = q;
    } else {
      if (a[q] == 1) {
        int x = s[i + 1][0] - 'A';
        int y = s[i + 1][1] - 'A';
        if (q == x || q == y) {
          ch = q;
        } else {
          ch = p;
        }
      } else {
        ch = p; 
      }
    }
    
    ++a[ch];
    --a[p ^ q ^ ch];
    ans += 'A' + ch;
  }
  return true;
}

int main() {
  cin >> n >> a[0] >> a[1] >> a[2];
  for (int i = 0; i < n; ++i) cin >> s[i];
  s[n] = "AB";
  if (ok()) {
    cout << "Yes" << endl;
    for (int i = 0; i < n; ++i) {
      cout << ans[i] << endl;
    }
  } else {
    cout << "No" << endl;
  }
}
### Prompt

Please formulate 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 n;
int a[3]; 
string s[100001];
string ans;

bool ok() {
  for (int i = 0; i < n; ++i) {
    int p = s[i][0] - 'A';
    int q = s[i][1] - 'A';
    if (a[p] == 0 && a[q] == 0) return false;
    int ch;
    if (a[p] == 0) {
      ch = p;
    } else if (a[q] == 0) {
      ch = q;
    } else {
      if (a[q] == 1) {
        int x = s[i + 1][0] - 'A';
        int y = s[i + 1][1] - 'A';
        if (q == x || q == y) {
          ch = q;
        } else {
          ch = p;
        }
      } else {
        ch = p; 
      }
    }
    
    ++a[ch];
    --a[p ^ q ^ ch];
    ans += 'A' + ch;
  }
  return true;
}

int main() {
  cin >> n >> a[0] >> a[1] >> a[2];
  for (int i = 0; i < n; ++i) cin >> s[i];
  s[n] = "AB";
  if (ok()) {
    cout << "Yes" << endl;
    for (int i = 0; i < n; ++i) {
      cout << ans[i] << endl;
    }
  } else {
    cout << "No" << endl;
  }
}
```