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>
#define rep(i, n) for (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
using Graph = vector<vector<int>>;
#define MOD 1000000007
#define MOD2 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;


string s[100010];
ll now[3];
ll after[3][100010]; // 今後の出現回数
vector<int> ans;

int main(){
    ll n, a, b, c;
    cin >> n >> a >> b >> c;

    rep(i, n) cin >> s[i];

    now[0] = a;
    now[1] = b;
    now[2] = c;
    bool flag = true;

    rrep(i, n){
        int p = s[i][0] - 'A';
        int q = s[i][1] - 'A';
        after[p][i] = after[p][i+1] + 1;
        after[q][i] = after[q][i+1] + 1;
        after[3-p-q][i] = after[3-p-q][i+1];
    }

    // rep(i, n){
    //     cout << after[0][i] << " " << after[1][i] << " " << after[2][i] << endl;
    // }

    rep(i, n){
        int p = s[i][0] - 'A';
        int q = s[i][1] - 'A';
        if (now[p] == 0 && now[q] == 0){
            flag = false;
            break;
        }
        else{
            if(now[p] > now[q]){
                now[p]--;
                now[q]++;
                ans.push_back(q);
            }
            else if(now[p] < now[q]){
                now[q]--;
                now[p]++;
                ans.push_back(p);
            }
            else{
                if (after[p][i+1] >= after[q][i+1]){
                    now[q]--;
                    now[p]++;
                    ans.push_back(p);
                }
                else{
                    now[p]--;
                    now[q]++;
                    ans.push_back(q);
                }
            }
        }
    }

    if (!flag) {
        cout << "No" << endl;
        return 0;
    }
    
    cout << "Yes" << endl;
    rep(i, n){
        cout << char(ans[i] + 'A') << endl;
    }



    return 0;
}