#include <bits/stdc++.h> 
#define ll long long
#define F first
#define FF first.first
#define FS first.second
#define S second
#define pb push_back
using namespace std;
const ll N=1000006, INF=1e16;
 
ll n, q=1, m, e[4], a[N], b[N], c[N], o[4], k, t, x, y, z, u[4], h, mx, mn, ans, Ans;
pair<ll,ll> p[N];
string s[N];
vector<int> v[N];
 
int main(){ios_base::sync_with_stdio(false), cin.tie(0);
   	
	cin>>n>>e[1]>>e[2]>>e[3];
	
	for (int i=1; i<=n; i++)
		cin>>s[i];
		
	for (int i=1; i<=n; i++){
		
		x=s[i][0]-'A'+1;
		y=s[i][1]-'A'+1;
		
		if(e[x]>e[y])swap(x, y);
		
		if(e[x]==e[y] && e[x]==1 && i<n && ((s[i+1][0]-'A'+1)==6-x-y || (s[i+1][1]-'A'+1)==6-x-y)){
			if((s[i+1][0]-'A'+1)==6-x-y){
				k=(s[i+1][1]-'A'+1);
				e[k]++;
				e[(x^y^k)]--;
				a[i]=k;
			}
			if((s[i+1][1]-'A'+1)==6-x-y){
				k=(s[i+1][0]-'A'+1);
				e[k]++;
				e[(x^y^k)]--;
				a[i]=k;
			}
		}
		if(!a[i]){
			e[x]++;
			e[y]--;
			
			a[i]=x;
		}
		if(e[y]<0){
			cout<<"No"; return 0;
		}
	}
	
	cout<<"Yes"<<endl;
	for (int i=1; i<=n; i++){
		cout<<char('A'+a[i]-1)<<endl;
	}
}
### Prompt

Create a solution in Cpp for 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 ll long long
#define F first
#define FF first.first
#define FS first.second
#define S second
#define pb push_back
using namespace std;
const ll N=1000006, INF=1e16;
 
ll n, q=1, m, e[4], a[N], b[N], c[N], o[4], k, t, x, y, z, u[4], h, mx, mn, ans, Ans;
pair<ll,ll> p[N];
string s[N];
vector<int> v[N];
 
int main(){ios_base::sync_with_stdio(false), cin.tie(0);
   	
	cin>>n>>e[1]>>e[2]>>e[3];
	
	for (int i=1; i<=n; i++)
		cin>>s[i];
		
	for (int i=1; i<=n; i++){
		
		x=s[i][0]-'A'+1;
		y=s[i][1]-'A'+1;
		
		if(e[x]>e[y])swap(x, y);
		
		if(e[x]==e[y] && e[x]==1 && i<n && ((s[i+1][0]-'A'+1)==6-x-y || (s[i+1][1]-'A'+1)==6-x-y)){
			if((s[i+1][0]-'A'+1)==6-x-y){
				k=(s[i+1][1]-'A'+1);
				e[k]++;
				e[(x^y^k)]--;
				a[i]=k;
			}
			if((s[i+1][1]-'A'+1)==6-x-y){
				k=(s[i+1][0]-'A'+1);
				e[k]++;
				e[(x^y^k)]--;
				a[i]=k;
			}
		}
		if(!a[i]){
			e[x]++;
			e[y]--;
			
			a[i]=x;
		}
		if(e[y]<0){
			cout<<"No"; return 0;
		}
	}
	
	cout<<"Yes"<<endl;
	for (int i=1; i<=n; i++){
		cout<<char('A'+a[i]-1)<<endl;
	}
}
```