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:
/*input
8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB
*/
#include "bits/stdc++.h"
using namespace std;

long long n, a, b, c;
string s[100005];

map<char, int> mp[100005], have;
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n >> have['A'] >> have['B'] >> have['C'];

	for(int i = 1; i <= n; ++i){
		cin >> s[i];
	}
	for(int i = n; i >= 0; --i){
		for(auto c: {'A', 'B', 'C'})
			mp[i][c] = count(s[i].begin(), s[i].end(), c) + mp[i][c];
	}
	vector<char> RES;
	for(int i = 1; i <= n; ++i){
		char c, d;
		string s = ::s[i];
		if(have[s[0]] + have[s[1]] == 0)
			return cout << "No\n", 0;
		else if(have[s[0]] == 0)
			c = s[1], d = s[0];
		else if(have[s[1]] == 0 || i == n)
			c = s[0], d = s[1];
		else {
			if(count(::s[i+1].begin(),::s[i+1].end(),s[0]))
				c = s[1], d = s[0];
			else
				c = s[0], d = s[1];
		}
		RES.push_back(d);
		--have[c];
		++have[d];
	}
	cout << "Yes\n";
	for(auto i: RES) cout << i << '\n';
}