#include <bits/stdc++.h>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

const int max_n = 101010;
int n;
vector<string> v;

string ans[max_n];

void dfs(int i, int a, int b, int c){
	if(a < 0 || b < 0 || c < 0) return;

	if(i == n){
		cout << "Yes" << "\n";
		rep(i, n) cout << ans[i] << "\n";
		exit(0);
	}

	if(v[i] == "AB"){
		ans[i] = "A";
		dfs(i+1, a+1, b-1, c);
		ans[i] = "B";
		dfs(i+1, a-1, b+1, c);
	}
	
	if(v[i] == "BC"){
		ans[i] = "B";
		dfs(i+1, a, b+1, c-1);
		ans[i] = "C";
		dfs(i+1, a, b-1, c+1);
	}

	if(v[i] == "AC"){
		ans[i] = "C";
		dfs(i+1, a-1, b, c+1);
		ans[i] = "A";
		dfs(i+1, a+1, b, c-1);
	}
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n;

	int a, b, c;
	cin >> a >> b >> c;

	v.resize(n);
	rep(i, n) cin >> v[i];

	dfs(0, a, b, c);
	cout << "No" << "\n";
}

### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

const int max_n = 101010;
int n;
vector<string> v;

string ans[max_n];

void dfs(int i, int a, int b, int c){
	if(a < 0 || b < 0 || c < 0) return;

	if(i == n){
		cout << "Yes" << "\n";
		rep(i, n) cout << ans[i] << "\n";
		exit(0);
	}

	if(v[i] == "AB"){
		ans[i] = "A";
		dfs(i+1, a+1, b-1, c);
		ans[i] = "B";
		dfs(i+1, a-1, b+1, c);
	}
	
	if(v[i] == "BC"){
		ans[i] = "B";
		dfs(i+1, a, b+1, c-1);
		ans[i] = "C";
		dfs(i+1, a, b-1, c+1);
	}

	if(v[i] == "AC"){
		ans[i] = "C";
		dfs(i+1, a-1, b, c+1);
		ans[i] = "A";
		dfs(i+1, a+1, b, c-1);
	}
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n;

	int a, b, c;
	cin >> a >> b >> c;

	v.resize(n);
	rep(i, n) cin >> v[i];

	dfs(0, a, b, c);
	cout << "No" << "\n";
}

```