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)

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

int main() {
    int N, A, B, C;
    cin >> N >> A >> B >> C;
    vector<string> S(N);
    vector< vector<int> > num(N + 1, vector<int>(3)); // num[i][j]: i回目までのjの登場回数
    FOR(i, 1, N) {
        cin >> S[i-1];
        if(S[i][0] == 'A') num[i][0] = num[i-1][0] + 1;
        else num[i][0] = num[i-1][0];
        if(S[i][0] == 'B' || S[i][1] == 'B') num[i][1] = num[i-1][1] + 1;
        else num[i][1] = num[i-1][1];
        if(S[i][1] == 'C') num[i][2] = num[i-1][2] + 1;
        else num[i][2] = num[i-1][2];
    }

    // FOR(i, 1, N) {
    //     cerr << num[i][0] << " " << num[i][1] << " " << num[i][2] << endl;
    //     if(A + B ) {
    //         cout << "No" << endl;
    //         return 0;
    //     }
    // }

    // 可能な場合
    vector<char> ans;
    vector<int> cur(3);
    cur[0] = A;
    cur[1] = B;
    cur[2] = C;
    REP(i, N) {
        if(S[i] == "AB")
            if(cur[0] > cur[1]) {
                ans.push_back('B');
                cur[0]--;
                cur[1]++;
            } else {
                ans.push_back('A');
                cur[0]++;
                cur[1]--;
            }
        else if(S[i] == "BC")
            if(cur[2] > cur[1]) {
                ans.push_back('B');
                cur[2]--;
                cur[1]++;
            } else {
                ans.push_back('C');
                cur[2]++;
                cur[1]--;
            }
        else
            if(cur[0] > cur[2]) {
                ans.push_back('C');
                cur[0]--;
                cur[2]++;
            } else {
                ans.push_back('A');
                cur[0]++;
                cur[2]--;
            }

        if(cur[0] < 0 || cur[1] < 0 || cur[2] < 0) {
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
    for(char x: ans) cout << x << endl;

    return 0;
}