#include<bits/stdc++.h>
#define M 1000000007
#define pi ld(acos(-1.0))
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define Rep(i,n) for(int i=1;i<(int)(n);i++)
#define repp(i,a,b)  for(int i=a;i<=(int)(b);i++)
#define Repp(i,a,b)  for(int i=a;i>=(int)(b);i--)
#define cs ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ii pair<int,int>
#define bye(x) {cout<<x;return 0;}
typedef long long ll;
typedef long double ld;
using namespace std;

int main(){
    cs
    int n,a,b,c;cin>>n;
    vector<int> v(3);
    vector<string> ins(n);
    vector<char> ans(n);
    for(auto &i:v) cin>>i;
    for(auto &i:ins) cin>>i;
    repp(i,0,n-1) {
        int t=ins[i][0]-'A',tt=ins[i][1]-'A';
        if(v[t]==0&&v[tt]==0) bye("No")
        if((v[t]>v[tt])||
        (v[t]==v[tt]&&i!=n-1&&(ins[i][1]==ins[i+1][0]
        ||ins[i][1]==ins[i+1][1]))) swap(t,tt);
        v[t]++;v[tt]--;
        ans[i]=char(t+'A');
    }
    cout<<"Yes\n";
    for(auto i:ans) cout<<i<<'\n';
}
### Prompt

Your task is to create 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>
#define M 1000000007
#define pi ld(acos(-1.0))
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define Rep(i,n) for(int i=1;i<(int)(n);i++)
#define repp(i,a,b)  for(int i=a;i<=(int)(b);i++)
#define Repp(i,a,b)  for(int i=a;i>=(int)(b);i--)
#define cs ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ii pair<int,int>
#define bye(x) {cout<<x;return 0;}
typedef long long ll;
typedef long double ld;
using namespace std;

int main(){
    cs
    int n,a,b,c;cin>>n;
    vector<int> v(3);
    vector<string> ins(n);
    vector<char> ans(n);
    for(auto &i:v) cin>>i;
    for(auto &i:ins) cin>>i;
    repp(i,0,n-1) {
        int t=ins[i][0]-'A',tt=ins[i][1]-'A';
        if(v[t]==0&&v[tt]==0) bye("No")
        if((v[t]>v[tt])||
        (v[t]==v[tt]&&i!=n-1&&(ins[i][1]==ins[i+1][0]
        ||ins[i][1]==ins[i+1][1]))) swap(t,tt);
        v[t]++;v[tt]--;
        ans[i]=char(t+'A');
    }
    cout<<"Yes\n";
    for(auto i:ans) cout<<i<<'\n';
}
```