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>

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vector<int>> vvi;
typedef vector<string> vs;
typedef unordered_set<int> si;
typedef pair<int, int> pp;
typedef vector<pp> vpp;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#define all(e) (e).begin(), (e).end()
#define pb push_back


template <typename T>
void printV(vector<T> A) {
    if (A.empty()) cerr << "EMTPY\n";
    else { 
        cerr << "[";
        for (int i = 0; i < A.size() -1 ; ++i) {
            cerr << A[i] << ", ";
        }
        if (A.size()) cerr << A.back() << "]\n";
    }
}


template <typename T>
void printVV(vector<vector<T>> A) {
    cerr << "[";
    for (auto v : A) {
        printV<T>(v);
    }
    cerr << "]";
}

template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
    cerr << name << ": " << arg1 << endl;
}

template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&& ...args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << " |";
    __f(comma + 1, args...);
}

ll n,a,b,c;

int main(int argc, char **argv) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> a >> b >> c;
    ll cnt[3] = {a,b,c};
    vector<string> q(n);
    for (int i = 0; i < n; ++i) {
        cin >> q[i];
    }
    //last one check
    vector<char> ans(n);
    for (int i = 0; i < n-1; ++i) {
        string& cur = q[i], nxt = q[i+1];
        if (cur == nxt) {
            // no care
            if (cnt[cur[0] - 'A'] > 0) {
                ans[i] = cur[1];
                cnt[cur[0] - 'A'] -= 1, cnt[cur[1] - 'A'] += 1;
            }
            else {
                ans[i] = cur[0];
                cnt[cur[1] - 'A'] -= 1, cnt[cur[0] - 'A'] += 1;
            }
            continue;
        }
        char common,weak;
        if (cur[0] == nxt[0] || cur[0] == nxt[1]) weak = cur[1], common = cur[0];
        else common = cur[1], weak = cur[0];
        if (cnt[weak-'A'] > 0) {
            cnt[weak-'A'] -= 1, cnt[common-'A'] += 1;
            ans[i] = common;
        }
        else {
            if (cnt[common-'A'] == 0) {
                cout << "No\n";
                return 0;
            }
            else  {
                cnt[weak-'A'] += 1, cnt[common-'A'] -= 1;
                ans[i] = weak;
            }
        }
    }

    string& lst = q[n-1];
    if (cnt[lst[0]-'A'] == 0 && cnt[lst[1] - 'A'] == 0) {
        cout << "No\n";
        return 0;
    }
    else {
        if (cnt[lst[0]-'A'] > cnt[lst[1]-'A']) ans[n-1] = lst[0];
        else ans[n-1] = lst[1];
    }


    cout << "Yes\n";
    for (int i = 0; i < n; ++i) {
        cout << ans[i] << '\n';
    }
    cout << endl;
    return 0;
}