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>

#ifdef DEBUG
#define debug(x) cout << #x << ": " << x << endl
#else
#define debug(x)
#endif

#define REP(i, n) for(int i = 0; i < (n); i++)
#define REPS(i, n) for(int i = 1; i <= (n); i++)
#define ALL(v) (v).begin(), (v).end()

#define pb push_back
#define mp make_pair
#define mt make_tuple

using namespace std;

using ll = long long;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;

const ll INF = 1e18;
const ll MOD = 1e9+7;

template<class T> inline bool chmax(T &a, const T &b) { if (a<b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, const T &b) { if (a>b) { a = b; return true;} return false; }


signed main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    ll N, A, B, C; cin >> N >> A >> B >> C;
    vs s(N);
    map<char, ll> n;
    n['A'] = A;
    n['B'] = B;
    n['C'] = C;
    REP(i, N)
    {
        cin >> s[i];
        if(s[i] == "AB")
        {
            n['A'] -= 1;
            n['B'] -= 1;
        }
        if(s[i] == "BC")
        {
            n['B'] -= 1;
            n['C'] -= 1;
        }
        if(s[i] == "AC")
        {
            n['A'] -= 1;
            n['C'] -= 1;
        }
    }

    vector<char> ans;
    REP(i, N)
    {
        if(s[i] == "AB")
        {
            if(n['A'] < n['B'])
            {
                n['A'] += 2;
                ans.pb('A');
            }
            else
            {
                n['B'] += 2;
                ans.pb('B');
            }
        }
        if(s[i] == "BC")
        {
            if(n['B'] < n['C'])
            {
                n['B'] += 2;
                ans.pb('B');
            }
            else
            {
                n['C'] += 2;
                ans.pb('C');
            }
        }
        if(s[i] == "AC")
        {
            if(n['A'] < n['C'])
            {
                n['A'] += 2;
                ans.pb('A');
            }
            else
            {
                n['C'] += 2;
                ans.pb('C');
            }
        }
    }

    if(n['A'] < 0 || n['B'] < 0 || n['C'] < 0)
    {
        cout << "No" << endl;
        return 0;
    }
    cout << "Yes" << endl;
    REP(i, N) cout << ans[i] << endl;
}