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: Runtime Error)

// g++ -std=c++11 a.cpp
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<unordered_map>
#include<utility>
#include<cmath>
#include<random>
#include<cstring>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#include<typeinfo>
#define loop(i,a,b) for(long long i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
#define FOR(i,a) for(auto i:a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
#define show1d(v) {rep(_,v.size())cout<<" "<<v[_];cout<<endl;}
#define show2d(v) {rep(__,v.size())show1d(v[__]);}
using namespace std;
//kaewasuretyuui
typedef long long ll;
#define int ll
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
#define mt make_tuple
typedef tuple<int,int,int> tp;
typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-9;
Def inf = sizeof(Def) == sizeof(long long) ? 2e18 : 1e9+10;
int dx[]={-1,0,1,0};
int dy[]={0,1,0,-1};
#define yes cout<<"Yes\n"
#define no cout<<"No\n"

//nCr mod m
#define MOD 998244353
#define M 2200000
vector<ll> fact;
bool h=false;
void init(){
	h=true;
	fact=vector<ll>(M);
	fact[0]=fact[1]=1;
	loop(i,2,M)fact[i]=fact[i-1]*i%MOD;
}
// a^b mod MOD
ll powmod(ll a,ll b,ll m=MOD){
	ll out=1;
	ll p=a%m;
	while(b){
		if(b&1)out=out*p%m;
		p=p*p%m;
		b>>=1;
	}
	return out;
}
//nCr
ll nCr(ll n,ll r,ll m=MOD){
	if(!h)init();
	if(n<0||r<0||n<r)return 1;//??????
	ll out=fact[n]*powmod(fact[r]*fact[n-r]%m,m-2,m)%m;
	return out;
}
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;
}

int n,t[3];
vs in,out;
void f(int a,int b){
	t[a]++;
	t[b]--;
}
int dfs(int c){
	int a=in[c][0]-'A',b=in[c][1]-'A';
//	cout<<c<<" "<<t[0]<<" "<<t[1]<<" "<<t[2]<<endl;
	if(t[0]<0||t[1]<0||t[2]<0)return 0;
	if(c==n)return 1;
	if(t[a]<t[b]){
		f(a,b);
		out[c]=a+'A';
		if(dfs(c+1))return 1;
		f(b,a);
	}
	{
		f(b,a);
		out[c]=b+'A';
		if(dfs(c+1))return 1;
		f(a,b);
	}
	{
		f(a,b);
		out[c]=a+'A';
		if(dfs(c+1))return 1;
		f(b,a);
	}
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin>>n>>t[0]>>t[1]>>t[2];
	in=out=vs(n);
	rep(i,n)cin>>in[i];

	dfs(0);
	yes;
	rep(i,n)cout<<out[i]<<endl;
}