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)

/*input
3 0 0 0
AB
BC
AC
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

#define rep(i,j,n) for(i=j;i<n;i++)
#define repi(i,j,n) for(i=j;i>n;i--)
#define inf 100001
#define M 1000000007
#define ll long long
#define ld long double
#define vll vector<long long>
#define vi vector<int>
#define pll pair<long long,long long>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define vpll vector<pair<long long,long long> >
#define pb push_back
#define pf pop_front
#define F first
#define S second
#define endl '\n'

using namespace std;
using namespace __gnu_pbds; 

template<typename T>
using ordered_set=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;


ll power(ll a,ll m,ll mod){
	ll ans=1;
	while(m){
		if(m%2){
			ans*=a;
			ans%=(mod);
		}
		a=(a*a)%(mod);
		m>>=1;
	}
	return ans;
}


void init(ll* a,ll n,ll val){
	for(ll i=0;i<n;i++)
		a[i]=val;
}



int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);	
	ll t=1,i,j;
	ll n,a1,b1,c1,k;
	cin>>n>>a1>>b1>>c1;
	string s[n];
	ll ab=0,bc=0,ac=0;
	rep(i,0,n){cin>>s[i];if(s[i]=="AB")ab++;else if(s[i]=="AC")ac++;else bc++;}
	ab=min(3ll,ab);
	ac=min(3ll,ac);
	bc=min(3ll,bc);
	rep(i,-ab,ab+1){
		rep(j,-ac,ac+1){
			rep(k,-bc,bc+1){
				ll a=a1,b=b1,c=c1;
				a+=i,b-=i,a-=j,c+=j,b+=k,c-=k;
				if(a>-1&&b>-1&&c>-1){
					cout<<"Yes\n";
					ab=i,ac=j,bc=k;
					rep(i,0,n){
						if(s[i]=="AB"){
							if(ab<0)ab++,cout<<"B\n";
							else ab--,cout<<"A\n";
						}else if(s[i]=="AC"){
							if(ac<0)ac++,cout<<"A\n";
							else ac--,cout<<"C\n";
						}else{
							if(bc<0)bc++,cout<<"C\n";
							else bc--,cout<<"B\n";
						}
					}
					return 0;
				}
			}
		}
	}
	cout<<"No\n";
	return 0;
}