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 int long long
using namespace std;

const int N = 1e5 + 5;

int n;
string s[N];
map<char, int> mp;
char ans[N];

signed main() {
	cin >> n >> mp['A'] >> mp['B'] >> mp['C'];
	for (int i = 1; i <= n; i++) {
		cin >> s[i];
	}
	for (int i = 1; i <= n; i++) {
		if (mp[s[i][0]] == 0 && mp[s[i][1]] == 0) {
			cout << "No" << endl;
			return 0;
		}
		if (mp[s[i][0]] > mp[s[i][1]]) {
			mp[s[i][0]]--;
			mp[s[i][1]]++;
			ans[i] = s[i][1];
		} else if (mp[s[i][0]] < mp[s[i][1]]) {
			mp[s[i][0]]++;
			mp[s[i][1]]--;
			ans[i] = s[i][0];
		} else if (mp[s[i][0]] == mp[s[i][1]]) {
			if (i == n || (s[i + 1][0] != s[i][0] && s[i + 1][1] != s[i][0])) {
				mp[s[i][0]]--;
			   mp[s[i][1]]++;
			   ans[i] = s[i][1];
			} else {
				mp[s[i][0]]++;
			   mp[s[i][1]]--;
			   ans[i] = s[i][0];
			}
		}
	}
	cout << "Yes" << endl;
	for (int i = 1; i <= n; i++) {
		cout << ans[i] << endl;
	}
	return 0;
}