#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int main() {
	int n; cin >> n;
	VI a(3); REP(i,3) cin >> a[i];
	vector<string> s(n);
	VVI p(3,VI(0));
	REP(i,n){
		cin >> s[i];
		p[s[i][0]-'A'].push_back(i);
		p[s[i][1]-'A'].push_back(i);
	}
	VI pm(3);
	REP(i,3) pm[i]=p[i].size();
	VI nowp(3,0);
	vector<char> ans;
	REP(i,n){
		int x=s[i][0]-'A', y=s[i][1]-'A';
		if(a[x]==0&&a[y]==0){
			cout << "No" << endl;
			return 0;
		}
		int lifex, lifey;
		if(nowp[x]+a[x]>=pm[x])
			lifex=INF;
		else
			lifex=p[x][nowp[x]+a[x]];
		if(nowp[y]+a[y]>=pm[y])
			lifey=INF;
		else
			lifey=p[y][nowp[y]+a[y]];
		if(lifex<lifey){
			ans.push_back(s[i][0]);
			a[x]++; a[y]--;
		}
		else{
			ans.push_back(s[i][1]);
			a[x]--; a[y]++;
		}
		nowp[x]++; nowp[y]++;
	}
	cout << "Yes" << endl;
	REP(i,n)
		cout << ans[i] << endl;
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int main() {
	int n; cin >> n;
	VI a(3); REP(i,3) cin >> a[i];
	vector<string> s(n);
	VVI p(3,VI(0));
	REP(i,n){
		cin >> s[i];
		p[s[i][0]-'A'].push_back(i);
		p[s[i][1]-'A'].push_back(i);
	}
	VI pm(3);
	REP(i,3) pm[i]=p[i].size();
	VI nowp(3,0);
	vector<char> ans;
	REP(i,n){
		int x=s[i][0]-'A', y=s[i][1]-'A';
		if(a[x]==0&&a[y]==0){
			cout << "No" << endl;
			return 0;
		}
		int lifex, lifey;
		if(nowp[x]+a[x]>=pm[x])
			lifex=INF;
		else
			lifex=p[x][nowp[x]+a[x]];
		if(nowp[y]+a[y]>=pm[y])
			lifey=INF;
		else
			lifey=p[y][nowp[y]+a[y]];
		if(lifex<lifey){
			ans.push_back(s[i][0]);
			a[x]++; a[y]--;
		}
		else{
			ans.push_back(s[i][1]);
			a[x]--; a[y]++;
		}
		nowp[x]++; nowp[y]++;
	}
	cout << "Yes" << endl;
	REP(i,n)
		cout << ans[i] << endl;
	return 0;
}
```