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;

const int N = 1e5 + 10;

int n, d[3];
int a[3][N], step[N];

int finddif(int x, int y) {
    for (int i = 0; i < 3; i++)
        if (a[i][x] && i != y)
            return i;
}

void solve(int p0, int p1, int p2) {
    int t[3];
    t[0] = d[0]; t[1] = d[1]; t[2] = d[2];
    for (int i = 0; i < n; i++) {
        if (a[p0][i]) {
            t[p0]++; t[finddif(i, p0)]--;
            step[i] = p0;
        } else {
            t[p1]++; t[finddif(i, p1)]--;
            step[i] = p1;
        }
        if (t[0] < 0) return;
        if (t[1] < 0) return;
        if (t[2] < 0) return;
    }
    puts("Yes");
    for (int i = 0; i < n; i++)
        cout << (char)(step[i] + 'A') << endl;
    exit(0);
}

void Try1() {
    int t[3];
    t[0] = d[0]; t[1] = d[1]; t[2] = d[2];
    for (int i = 0; i < n; i++) {
        int x = rand() % 2 + 1;
        if (t[a[!x][i]] == 0) x ^= 1;
        t[a[x][i]]++; t[a[!x][i]]--;
        step[i] = x;
        if (t[0] < 0) return;
        if (t[1] < 0) return;
        if (t[2] < 0) return;
    }
    puts("Yes");
    for (int i = 0; i < n; i++)
        cout << (char)(step[i] + 'A') << endl;
    exit(0);
}

void Try2() {
    int t[3];
    t[0] = d[0]; t[1] = d[1]; t[2] = d[2];
    for (int i = 0; i < n; i++) {
        int x = 0;
        if (t[a[x][i]] < t[a[!x][i]]) x ^= 1;
        t[a[x][i]]++; t[a[!x][i]]--;
        step[i] = x;
        if (t[0] < 0) return;
        if (t[1] < 0) return;
        if (t[2] < 0) return;
    }
    puts("Yes");
    for (int i = 0; i < n; i++)
        cout << (char)(step[i] + 'A') << endl;
    exit(0);
}

int main() {
    if (fopen("A.inp", "r"))
    freopen("A.inp", "r", stdin),
    freopen("A.out", "w", stdout);
    cin >> n >> d[0] >> d[1] >> d[2];
    for (int i = 0; i < n; i++) {
        string str; cin >> str;
        a[str[0] - 'A'][i] = a[str[1] - 'A'][i] = 1;
    }
    solve(0, 1, 2);
    solve(0, 2, 1);
    solve(1, 0, 2);
    solve(1, 2, 0);
    solve(2, 0, 1);
    solve(2, 1, 0);
//    for (int i = 0; i < 200; i++)
//        Try1();
//    Try2();
    puts("No");
}