Problem: 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.

Examples

Input

2 1 3 0
AB
AC


Output

Yes
A
C


Input

3 1 0 0
AB
BC
AB


Output

No


Input

1 0 9 0
AC


Output

No


Input

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB


Output

Yes
C
B
B
C
C
B
A
A

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	vector<int> d(3);
	cin >> n;
	rep(i,3) cin >> d[i];
	vector<P> q;
	rep(i,n){
		string s;
		cin >> s;
		int a = 0, b = 1;
		if(s == "AB") a = 0, b = 1;
		if(s == "BC") a = 1, b = 2;
		if(s == "AC") a = 0, b = 2;
		q.emplace_back(a, b);
	}
	string ans;
	auto add = [&](int a, int b){
		d[a]++;
		d[b]--;
		ans += 'A' + a;
	};
	q.emplace_back(0, 1);
	rep(i,n){
		int a = q[i].first;
		int b = q[i].second;
		if(!d[a] && !d[b]){
			cout << "No" << endl;
			return 0;
		}
		if(!d[a]){
			add(a, b);
		}
		else if(!d[b]){
			add(b, a);
		}
		else{
			int na = q[i+1].first;
			int nb = q[i+1].second;
			if(a == na || a == nb){
				add(a, b);
			}
			else{
				add(b, a);
			}
		}
	}
	cout << "Yes" << endl;
	rep(i,n) cout << ans[i] << endl;
	return 0;
}