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 PREP(i, s, x) for(ll i = (s); i < (x); i ++)
#define MREP(i, s, x) for(ll i = (s); i >= (x); i --)
#define REP(i, x) PREP(i, 0, x)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return 1;}return 0;}
template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return 1;}return 0;}

constexpr ll INF = (1ll << 60);
constexpr ll MOD = 1000000007ll;


// べき乗を計算　指数を計算
ll llpow(ll a, ll n){
    ll res = 1;
    while(n > 0){
        if(n & 1)
            res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

ll f(ll a){
    return a * a * a * a * a;
}


int main(){
    ll n, a, b, c;
    cin >> n >> a >> b >> c;
    string ans = "";
    REP(i, n){
        string s;
        cin >> s;
        if(s == "AB"){
            if(a == 0 && b == 0){
                cout << "No" << endl;
                return 0;
            }
            ll ap = abs(a + 1 - (b - 1));
            ll bp = abs(a - 1 - b + 1);
            if(ap > bp){
                b++;
                a--;
                ans += "B";
            }else{
                a++;
                b--;
                ans += "A";
            }
        }else if(s == "AC"){
            if(a == 0 && c == 0){
                cout << "No" << endl;
                return 0;
            }
            ll ap = abs(a + 1 - (c - 1));
            ll cp = abs(a - 1 - c + 1);
            if(ap > cp){
                c++;
                a--;
                ans += "C";
            }else{
                a++;
                c--;
                ans += "A";
            }
        }else{
            if(c == 0 && b == 0){
                cout << "No" << endl;
                return 0;
            }
            ll cp = abs(c + 1 - (b - 1));
            ll bp = abs(c - 1 - b + 1);
            if(cp > bp){
                b++;
                c--;
                ans += "B";
            }else{
                c++;
                b--;
                ans += "C";
            }
        }
        if(a < 0 || b < 0 || c < 0){
            return -1;
        }
    }
    cout << "Yes" << endl;
    REP(i, n){
        cout << ans[i] << endl;
    }
    return 0;
}