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;

int n,tp,cnt[256];
char ans[100005];
string s[100005];

int main()
{
    cin>>n>>cnt['A']>>cnt['B']>>cnt['C'];
    for(int i=1;i<=n;i++)
    {
        cin>>s[i];
        if(s[i][0]>s[i][1]) swap(s[i][0],s[i][1]);
    }
    for(int i=1;i<=n;i++)
    {
        if(cnt[s[i][0]]+cnt[s[i][1]]==0) return printf("No"),0;
        if(cnt[s[i][0]]==0||cnt[s[i][1]]==0)
        {
            if(cnt[s[i][1]]==0) swap(s[i][0],s[i][1]);
            ans[i]=s[i][0];
            cnt[s[i][0]]++,cnt[s[i][1]]--;
            continue;
        }
        if(i==n||(s[i+1][0]==s[i][0])+(s[i+1][1]==s[i][0])==1)
            ans[i]=s[i][0],cnt[s[i][0]]++,cnt[s[i][1]]--;
        else
            ans[i]=s[i][1],cnt[s[i][1]]++,cnt[s[i][0]]--;
    }
    printf("Yes\n");
    for(int i=1;i<=n;i++) printf("%c\n",ans[i]);
    return 0;
}