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 <iostream>
#include <string>
#include <stack>
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

int n, a[3];
string s[100000];

int next(int n) { return (n + 1) % 3; }
int prev(int n) { return (n - 1 + 3) % 3; }

signed main() {
	int i, j;
	
	cin >> n >> a[0] >> a[1] >> a[2];
	rep(i, n) {
		cin >> s[i];
		if (s[i] == "AC") { s[i] = "CA"; }
	}
	
	stack<int> stk[6];	//"AB", "BC", "CA" | "BA", "CB", "AC"
	rep(i, n) {
		a[s[i][0] - 'A']++;
		a[s[i][1] - 'A']--;
		stk[s[i][0] - 'A'].push(i);
		
		int t = s[i][0] - 'A';
		rep(j, 3) {
			while (!stk[t].empty() && a[next(t)] < 0) {
				int id = stk[t].top(); stk[t].pop();
				a[t] -= 2;
				a[next(t)] += 2;
				stk[t + 3].push(id);
			}
			t = prev(t);
		}
		if (a[0] < 0 || a[1] < 0 || a[2] < 0) { cout << "No" << endl; return 0; }
	}
	
	string coms[6] = {"AB", "BC", "CA", "BA", "CB", "AC"};
	static string ans[100000];
	
	rep(i, 6) {
		while (!stk[i].empty()) {
			int id = stk[i].top();
			stk[i].pop();
			ans[id] = coms[i];
		}
	}
	
	cout << "Yes" << endl;
	rep(i, n) cout << ans[i][0] << endl;
	return 0;
}