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>
using namespace std;

using ll = long long;

void NO () {
	cout << "No";
	exit (0);
}

void YES (string ans) {
	cout << "Yes\n" << ans;
	exit (0);
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, sum = 0, askiiA = 'A';
	cin >> n;
	vector <string> a(n);
	int c[3+askiiA];
	for (int i=askiiA; i<3+askiiA; i++) {
		cin >> c[i];
		sum += c[i];
	}
	if (!sum) NO();
	for (int i=0; i<n; i++) {
		cin >> a[i];
	}
	string ans = "";
	for (int i=0; i<n; i++) {
		string cur = a[i];
		if (!c[cur[0]] && !c[cur[1]]) NO();
		if (i == n-1) {
			if (c[cur[0]]) {
				ans += cur[1];
			} else {
				ans += cur[0];
			}
			break;
		}
		if (!c[cur[0]] && c[cur[1]]) {
			ans += cur[0];
			c[cur[0]] ++;
			c[cur[1]] --;
		} else if (c[cur[0]] && !c[cur[1]]) {
			ans += cur[1];
			c[cur[1]] ++;
			c[cur[0]] --;
		} else {
			string next = a[i+1];
			if (next[0] == cur[0] || next[1] == cur[0]) {
				ans += cur[0];
				c[cur[0]] ++;
				c[cur[1]] --;
			} else {
				ans += cur[1];
				c[cur[1]] ++;
				c[cur[0]] --;
			}
		}
		ans += "\n";
	}
	YES(ans);
	return 0;
}