Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
#include<complex>

using namespace std;

#define ll long long
#define ld long double
#define INF 1e9
#define LINF (ll)INF*INF
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(int i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

#define int ll //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

const double PI=acos(-1);
const double EPS=1e-9; //うまくいかなかったらゆるめる

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef pair<int,pii> pip;
typedef vector<pii> vp;
typedef vector<vi> vvi;

int gcd(int a, int b){if(b==0) return a;return gcd(b,a%b);}
int lcm(int a, int b){return a/gcd(a,b)*b;}


signed main(void) {
    int n,a,b,c;
    cin >> n >> a >> b >> c;
    string ans;
    rep(i,n){
        string s;
        cin >> s;
        if(s == "AB"){
            if(a == 0 && b == 0){
                cout << "No" << endl;
                return 0;
            }
            if(a < b)a++,b--,ans+='A';
            else a--,b++,ans+='B';
        }
        if(s == "AC"){
            if(a == 0 && c == 0){
                cout << "No" << endl;
                return 0;
            }
            if(a < c)a++,c--,ans+='A';
            else a--,c++,ans+='C';
        }
        if(s == "BC"){
            if(b == 0 && c == 0){
                cout << "No" << endl;
                return 0;
            }
            if(b < c)b++,c--,ans+='B';
            else b--,c++,ans+='C';
        }
    }
    cout << "Yes" << endl;
    rep(i,n)cout << ans[i] << endl;
}