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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using pint = pair<int, int>;
using ll = long long;
using ull = unsigned long long;
using vll = vector<long long>;
using pll = pair<ll, ll>;
#define FOR(i, begin, end) \
    for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) \
    for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define VREP(s, ite) for (auto ite = s.begin(); ite != s.end(); ++ite)
#define FI first
#define SE second
#define ALL(v) v.begin(), v.end()
//#define endl "\n"
#define ciosup  \
    cin.tie(0); \
    ios::sync_with_stdio(false);
#define eb emplace_back
#define vint vector<int>
constexpr ll INF = 1e15 + 7LL;
constexpr ll MOD = 1e9 + 7LL;

template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (int i = 0; i < v.size(); ++i) {
        is >> v[i];
    }
    return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    for (int i = 0; i < v.size() - 1; ++i) {
        os << v[i] << " ";
    }
    if (v.size() > 0) {
        os << v[v.size() - 1] << endl;
    }
    return os;
}

int main() {
    ll n,a,b,c;
    cin >> n >> a >> b >> c;
    vll nums{a, b, c};
    map<char, int> inds{{'A', 0}, {'B', 1}, {'C', 2}};
    vector<string> S(n);
    REP(i, n) {
        cin >> S[i];
    }
    vector<char> ans;
    REP(i, n) {
        string now = S[i];
        if (i == n - 1){
            if (nums[inds[now[0]]] == 0 && nums[inds[now[1]]] == 0){
                cout << "No" << endl;
                return 0;
            } else if (nums[inds[now[0]]] > 0) {
                ans.push_back(now[1]);
            }else{
                ans.push_back(now[0]);
            }
        } else {
            if (nums[inds[now[0]]] == 0 && nums[inds[now[1]]] == 0) {
                cout << "No" << endl;
                return 0;
            } else if (nums[inds[now[0]]] == 0) {
                ans.push_back(now[0]);
                nums[inds[now[0]]]++;
                nums[inds[now[1]]]--;
            } else if (nums[inds[now[1]]] == 0) {
                ans.push_back(now[1]);
                nums[inds[now[1]]]++;
                nums[inds[now[0]]]--;
            } else if (nums[inds[now[0]]] == 1 && nums[inds[now[1]]] > 1) {
                ans.push_back(now[0]);
                nums[inds[now[0]]]++;
                nums[inds[now[1]]]--;
            } else if (nums[inds[now[0]]] > 1 && nums[inds[now[1]]] == 1) {
                ans.push_back(now[1]);
                nums[inds[now[1]]]++;
                nums[inds[now[0]]]--;
            } else if (nums[inds[now[0]]] == 1 && nums[inds[now[1]]] == 1){
                string next = S[i+1];
                if (next[0] == now[0] || next[1] == now[0]) {
                    ans.push_back(now[0]);
                    nums[inds[now[0]]]++;
                    nums[inds[now[1]]]--;
                } else {
                    ans.push_back(now[1]);
                    nums[inds[now[1]]]++;
                    nums[inds[now[0]]]--;
                }
            } else {
                ans.push_back(now[0]);
                nums[inds[now[0]]]++;
                nums[inds[now[1]]]--;
            }
        }
    }
    cout << "Yes" << endl;
    REP(i, n) { cout << ans[i] << endl; }
}