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<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<assert.h>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=100000+100;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 998244353
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

int n,a[100100],b[5],m;
vector<pair<string,int> >ask;
string s[100100],ans="";

int work(string s,string t)
{
    b[0]=b[1]=b[2]=0;
    b[s[0]-'A']++;
    b[s[1]-'A']++;
    b[t[0]-'A']++;
    b[t[1]-'A']++;
    rep(i,0,2) cout << b[i] << " ";cout << endl;
    rep(i,0,2)
        if (b[i]==2) return i;
}

int main()
{
    n=read();
    rep(i,0,2) a[i]=read();
    rep(i,1,n) cin >> s[i];
    int c=1;
    //cout << work("BC","AB") << endl;
    rep(i,1,n)
    {
        if ((i==n) || (s[i]!=s[i+1]))
        {
            ask.pb(mkp(s[i],c));
            c=1;
        }
        else c++;
    }
    m=ask.size();
    rep(i,0,m-1)
    {
        string op=ask[i].fir;int c=ask[i].sec;
        assert(c>0);
        int x=op[0]-'A',y=op[1]-'A';
        if ((!a[x]) && (!a[y]))
        {
            puts("No");
            return 0;
        }
        else if ((a[x]==1) && (a[y]==1) && (i+1<m))
        {
            string nxtop=ask[i+1].fir;
            int s=work(nxtop,op),t=x+y-s;
            //cout << s << " " << t << endl;
            rep(j,1,c)
            {
                if (j&1) ans=ans+(char)('A'+s);else ans=ans+(char)('A'+t);
            }
            if (c&1) {a[s]++;a[t]--;}
        }
        else
        {
            if (a[x]<a[y]) swap(x,y);
            rep(j,1,c)
                if (j&1) ans=ans+(char)('A'+y);else ans=ans+(char)('A'+x);
            if (c&1) {a[x]--;a[y]++;}
        }
    }
    puts("Yes");
    rep(i,0,n-1) {putchar(ans[i]);puts("");}
    return 0;
}