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;
typedef long long ll;
#define REP(i, n) for (int i=0; i<n; i++)
#define println(v) cout << v << "\n";

int main()
{
	int N;
	cin >> N;
	vector<ll> var(256);
	for (char c='A'; c<='C'; c++) cin >> var[c];
	vector<string> s(N);
	REP(i, N) cin >> s[i];

	vector<char> ans(N);
	REP(i, N) {
		char c1 = s[i][0];
		char c2 = s[i][1];
		char cM = (var[c1] >= var[c2] ? c1 : c2);
		char cm = c1 ^ c2 ^ cM;
		if (var[c1] == var[c2] && i+1 < N && (s[i+1][0] == c1 || s[i+1][1] == c1)) {
			swap(cM, cm);
		}
		if (var[cM] == 0) {
			println("No");
			return 0;
		}
		var[cM]--;
		var[cm]++;
		ans[i] = cm;
	}

	println("Yes");
	REP(i, N) {
		println(ans[i]);
	}

	return 0;
}