#include <bits/stdc++.h>
#define all(x) begin(x), end(x)
#define pb push_back
#define int long long

using namespace std;
const int MOD = 1e9+7;

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	
	int n, a, b, c; cin >> n >> a >> b >> c;
	map<char, int> ct = {{'A', a}, {'B', b}, {'C', c}};
	vector<string> in(n);
	for (auto &s : in)
		cin >> s;
	in.pb("AB");

	string ans = "";
	for (int i = 0; i < n; ++i) {
		string s = in[i];
		int ch = 0; // Which character is being decremented?
		if (ct[s[0]] > ct[s[1]]) ch = 0; //  First
		else if (ct[s[1]] > ct[s[0]]) ch = 1; // Second
		else if (in[i+1][0] == s[0] or in[i+1][1] == s[0]) ch = 1;
		else ch = 0;
		--ct[s[ch]], ++ct[s[ch^1]];
		if (ct['A'] < 0 or ct['B'] < 0 or ct['C'] < 0) {ans = "-"; break;}
		else ans += s[ch^1], ans += "\n";
	}
	if (ans[0] == '-') cout << "No\n";
	else cout << "Yes\n" << ans;
}
### Prompt

In CPP, your task is to solve 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 all(x) begin(x), end(x)
#define pb push_back
#define int long long

using namespace std;
const int MOD = 1e9+7;

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	
	int n, a, b, c; cin >> n >> a >> b >> c;
	map<char, int> ct = {{'A', a}, {'B', b}, {'C', c}};
	vector<string> in(n);
	for (auto &s : in)
		cin >> s;
	in.pb("AB");

	string ans = "";
	for (int i = 0; i < n; ++i) {
		string s = in[i];
		int ch = 0; // Which character is being decremented?
		if (ct[s[0]] > ct[s[1]]) ch = 0; //  First
		else if (ct[s[1]] > ct[s[0]]) ch = 1; // Second
		else if (in[i+1][0] == s[0] or in[i+1][1] == s[0]) ch = 1;
		else ch = 0;
		--ct[s[ch]], ++ct[s[ch^1]];
		if (ct['A'] < 0 or ct['B'] < 0 or ct['C'] < 0) {ans = "-"; break;}
		else ans += s[ch^1], ans += "\n";
	}
	if (ans[0] == '-') cout << "No\n";
	else cout << "Yes\n" << ans;
}
```