#include<bits/stdc++.h>
#define int long long
#define ld long double
#define inf 0x7f7f7f7f7f7f7f7f 
#define maxn 1000010
#define ri register int
#define il inline
#define mk make_pair
#define pb push_back
#define fi first
#define se second
#define mod 998244353
#define db(x) cout << x << endl; 
#define eps 0.0001
using namespace std;
char ans[maxn];
string str[maxn];
int n;
void dfs(int i,int a,int b,int c) 
{
    if(i>n) 
	{
       	cout << "Yes\n";
        for(int i=1;i<=n;i++) 
			cout << ans[i] << endl;
        exit(0);
    }
    if(str[i]=="AB") 
	{
        if(a) ans[i]='B',dfs(i+1,a-1,b+1,c);
        if(b) ans[i]='A',dfs(i+1,a+1,b-1,c);
    }
    else if(str[i]=="AC") 
	{
        if(a) ans[i]='C',dfs(i+1,a-1,b,c+1);
        if(c) ans[i]='A',dfs(i+1,a+1,b,c-1);
    }
    else 
	{
        if(b) ans[i]='C',dfs(i+1,a,b-1,c+1);
        if(c) ans[i]='B',dfs(i+1,a,b+1,c-1);
    }
}
signed main()
{
	ios::sync_with_stdio(0);
	int a,b,c;
	cin >> n>> a >> b >> c;
	for(int i=1;i<=n;i++)
		cin >> str[i];
	dfs(1,a,b,c);
	cout << "No";
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 int long long
#define ld long double
#define inf 0x7f7f7f7f7f7f7f7f 
#define maxn 1000010
#define ri register int
#define il inline
#define mk make_pair
#define pb push_back
#define fi first
#define se second
#define mod 998244353
#define db(x) cout << x << endl; 
#define eps 0.0001
using namespace std;
char ans[maxn];
string str[maxn];
int n;
void dfs(int i,int a,int b,int c) 
{
    if(i>n) 
	{
       	cout << "Yes\n";
        for(int i=1;i<=n;i++) 
			cout << ans[i] << endl;
        exit(0);
    }
    if(str[i]=="AB") 
	{
        if(a) ans[i]='B',dfs(i+1,a-1,b+1,c);
        if(b) ans[i]='A',dfs(i+1,a+1,b-1,c);
    }
    else if(str[i]=="AC") 
	{
        if(a) ans[i]='C',dfs(i+1,a-1,b,c+1);
        if(c) ans[i]='A',dfs(i+1,a+1,b,c-1);
    }
    else 
	{
        if(b) ans[i]='C',dfs(i+1,a,b-1,c+1);
        if(c) ans[i]='B',dfs(i+1,a,b+1,c-1);
    }
}
signed main()
{
	ios::sync_with_stdio(0);
	int a,b,c;
	cin >> n>> a >> b >> c;
	for(int i=1;i<=n;i++)
		cin >> str[i];
	dfs(1,a,b,c);
	cout << "No";
	return 0;
}
```