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;

const int N=100020;
char ans[N];
char op[N][4];
int suf[N][3];
int n,a[3]={0};

void dfs(int dep)
{
    if (dep>n) {
        puts("Yes");
        for (int i=1;i<=n;i++) {
            printf("%c\n",ans[i]);
        }
        exit(0);
    }
    int b[3]={a[0],a[1],a[2]};
    for (int i=dep;i<=n+1;i++) {
        if (i>n) dfs(i);
        int x=op[i][0]-'A';
        int y=op[i][1]-'A';
        if (a[x]==0 && a[y]==0) return;
        if (a[x]<a[y]) {
            ans[i]=x+'A';
            a[x]++;
            a[y]--;
        } else if (a[x]>a[y]) {
            ans[i]=y+'A';
            a[y]++;
            a[x]--;
        } else {
            ans[i]=x+'A';
            a[x]++;
            a[y]--;
            dfs(i+1);
            ans[i]=y+'A';
            a[x]-=2;
            a[y]+=2;
            dfs(i+1);
            a[0]=b[0];
            a[1]=b[1];
            a[2]=b[2];
            return;
        }
    }
}

int main()
{
    scanf("%d%d%d%d",&n,a,a+1,a+2);
    for (int i=1;i<=n;i++) {
        scanf("%s",op[i]);
    }
    dfs(1);
    puts("No");
    return 0;
}