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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define pp pair<ll,ll>
#define ppp pair<ll,pp >
#define trace(...) cout<<"Line:"<<__LINE__<<" "; __f(#__VA_ARGS__, __VA_ARGS__)
template<typename Arg1>
void __f(const char* name, Arg1&& arg1) {
	cout<<name<<" : "<<arg1<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma=strchr(names+1,',');
	cout.write(names,comma-names)<<" : "<<arg1<<" | ";
	__f(comma+1,args...);
}
typedef tree<pp,null_type,less<pp>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
#define ld long double
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define rep(i,a,b) for( ll i = a ; i < b ; i+=1 )
#define sz(v) (ll)(v.size())
#define hell (ll)998244353
#define slld(x) scanf("%lld",&x)
ll t=1, tt;
const ll N=100005, inf=1e18;
void solve(){
	ll n, a, b, c;
	string s;
	cin>>n>>a>>b>>c;
	ll f=1;
	vector<char> ans;
	rep(i,0,n){
		cin>>s;
		if(s=="AB"){
			if(max(a,b)<=0)f=0;
			else{
				if(a<=b){
					a++;
					b--;
					ans.pb('A');
				}
				else{
					b++;
					a--;
					ans.pb('B');
				}
			}
		}
		else if(s=="BC"){
			if(max(c,b)<=0)f=0;
			else{
				if(c<=b){
					c++;
					b--;
					ans.pb('C');
				}
				else{
					b++;
					c--;
					ans.pb('B');
				}
			}
		}
		else{
			if(max(a,c)<=0)f=0;
			else{
				if(a<=c){
					a++;
					c--;
					ans.pb('A');
				}
				else{
					c++;
					a--;
					ans.pb('C');
				}
			}
		}
	}
	if(f==0){
		cout<<"No";
	}else{
		cout<<"Yes\n";
		for(auto u : ans)cout<<u<<"\n";
	}
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
   	//cin>>t;
   	tt=t;
    while(t--){
        solve();
        if(t)cout<<"\n";
    }
    return 0 ;
}