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: Accepted)

#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))

using namespace std;
typedef pair<int, int> P;

llint n, a[3];
string s[100005];
string ans;

llint get(string s)
{
	if(s == "AB") return 0;
	if(s == "BC") return 1;
	if(s == "AC") return 2;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	llint sum = 0;
	for(int i = 0; i < 3; i++) cin >> a[i], sum += a[i];
	for(int i = 1; i <= n; i++) cin >> s[i];
	
	for(int i = 1; i <= n; i++){
		llint x = get(s[i]), y = (x+1)%3, p, q;
		if(a[x] >= 2) p = x;
		else if(a[y] >= 2) p = y;
		else if(a[x] == 1 && a[y] == 1){
			if(i == n || get(s[i+1]) == y) p = x;
			else p = y;
		}
		else if(a[x] == 1) p = x;
		else if(a[y] == 1) p = y;
		else{
			cout << "No" << endl;
			return 0;
		}
		q = x;
		if(p == x) q = y;
		
		a[p]--, a[q]++;
		ans += "ABC"[q];
	}
	
	cout << "Yes" << endl;
	for(int i = 0; i < ans.size(); i++) cout << ans[i] << endl;
	
	return 0;
}