#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
vector<char> ans;
vector<string> s(100010);
void solve(ll& a, ll& b, char c) {
	a++;
	b--;
	ans.push_back(c);
}
bool move(ll& a, ll& b, string c, string s1, string s2) {
	if (a == 0 and b == 0) return false;
	else if (a == 1 and b == 1) s1 == s2 ? solve(a, b, c[0]) : solve(b, a, c[1]);
	else a < b ? solve(a, b, c[0]) : solve(b, a, c[1]);
	return true;
}
int main() {
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	REP(i, n) cin >> s[i];
	bool ok = true;
	REP(i, n) {
		if (s[i] == "AB") if (!move(a, b, "AB", "AC", s[i + 1])) ok = false;
		if (s[i] == "AC") if (!move(a, c, "AC", "AB", s[i + 1])) ok = false;
		if (s[i] == "BC") if (!move(b, c, "BC", "AB", s[i + 1])) ok = false;
	}
	if (!ok) {
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	REP(i, n) cout << ans[i] << endl;
}


### Prompt

Your challenge is to write 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>
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
vector<char> ans;
vector<string> s(100010);
void solve(ll& a, ll& b, char c) {
	a++;
	b--;
	ans.push_back(c);
}
bool move(ll& a, ll& b, string c, string s1, string s2) {
	if (a == 0 and b == 0) return false;
	else if (a == 1 and b == 1) s1 == s2 ? solve(a, b, c[0]) : solve(b, a, c[1]);
	else a < b ? solve(a, b, c[0]) : solve(b, a, c[1]);
	return true;
}
int main() {
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	REP(i, n) cin >> s[i];
	bool ok = true;
	REP(i, n) {
		if (s[i] == "AB") if (!move(a, b, "AB", "AC", s[i + 1])) ok = false;
		if (s[i] == "AC") if (!move(a, c, "AC", "AB", s[i + 1])) ok = false;
		if (s[i] == "BC") if (!move(b, c, "BC", "AB", s[i + 1])) ok = false;
	}
	if (!ok) {
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	REP(i, n) cout << ans[i] << endl;
}


```