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<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<a>>
#define UF  UnionFind 
#define Pint pair<int,int>
#define keta(a)  fixed << setprecision(a)
constexpr auto INF = 100000000000;
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！
struct edge { int to, cost; };



int main() {
	int n;
	cin >> n;
	llint a[3];
	rep(i, 0, 3) {
		cin >> a[i];
	}
	hei(string) s(n);
	rep(i, 0, n) {
		cin >> s[i];
	}
	
	string ans = "";
	rep(i, 0, n) {
		if (a[s[i][1] - 65] == 0 && a[s[i][0] - 65] == 0) {
		
			cout << "No" << endl;
			return 0;
		}
		if (a[s[i][1] - 65] == a[s[i][0] - 65] && a[s[i][0] - 65] == 1&&i!=n-1) {
			if (s[i][1] == s[i + 1][0] || s[i][1] == s[i + 1][0]) {
				a[s[i][1] - 65]++;
				a[s[i][0] - 65]--;
				ans += s[i][1];
			}
			else if(s[i][0]==s[i+1][0]||s[i][0]==s[i+1][1]) {
				a[s[i][1] - 65]--;
				a[s[i][0] - 65]++;
				ans += s[i][0];
			}
		}
		else {
			if (a[s[i][1] - 65] > a[s[i][0] - 65]) {
				a[s[i][1] - 65]--;
				a[s[i][0] - 65]++;
				ans += s[i][0];
			}
			else if(a[s[i][1]-65]<a[s[i][0]-65]){
				a[s[i][1] - 65]++;
				a[s[i][0] - 65]--;
				ans += s[i][1];
			}
		}
	}
	if (0<a[0] || 0<a[1] || 0<a[2]) {
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	rep(i, 0, n)cout << ans[i] << endl;
	return 0;
}