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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
long long n;
string s[100009];
char ans[100009];
map <char, long long> mp;

bool dfs(long long idx){
  if(idx==n)return true;
  if(mp[s[idx][0]]==0 && mp[s[idx][1]]==0)return false;
  if(mp[s[idx][0]]==mp[s[idx][1]]){
    bool ret;
    mp[s[idx][0]]++;
    mp[s[idx][1]]--;
    if(dfs(idx+1)){
      ans[idx] = s[idx][0];
      return true;
    }
    mp[s[idx][0]]-=2;
    mp[s[idx][1]]+=2;
    if(dfs(idx+1)){
      ans[idx] = s[idx][1];
      return true;
    }
    mp[s[idx][0]]--;
    mp[s[idx][1]]++;
    return false;
  }
  else if(mp[s[idx][0]]>mp[s[idx][1]]){
    mp[s[idx][0]]--;
    mp[s[idx][1]]++;
    ans[idx] = s[idx][1];
    return dfs(idx+1);
  }
  else{
    mp[s[idx][0]]++;
    mp[s[idx][1]]--;
    ans[idx] = s[idx][0];
    return dfs(idx+1);
  }
}

int main(){
  long long a, b ,c;
  cin >> n >> a >> b >> c;
  for(int i=0;i<n;i++)cin >> s[i];
  mp['A'] = a;
  mp['B'] = b;
  mp['C'] = c;
  if(dfs(0)){
    cout << "Yes" << endl;
    for(int i=0;i<n;i++)cout << ans[i] << endl;
  }
  else cout << "No" << endl;
}