#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using P=pair<int,int>;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,d[3];cin>>n;
    rep(i,3)cin>>d[i];
    vector<P> q;
    rep(i,n){
        string ss;cin>>ss;
        int a,b;
        if(ss=="AB")a=0,b=1;
        else if(ss=="BC")a=1,b=2;
        else a=0,b=2;
        q.push_back({a,b});
    }
    q.push_back({0,1});
    string ans;
    auto add=[&](int a,int b){
        d[a]++;d[b]--;
        ans+='A'+a;
    };
    rep(i,n){
        int a=q[i].first;
        int b=q[i].second;
        if(!d[a]&&!d[b]){
            cout<<"No"<<endl;
            return 0;
        }
        else if(!d[a])add(a,b);
        else if(!d[b])add(b,a);
        else {
            int na=q[i+1].first;
            int nb=q[i+1].second;
            if(a==na||a==nb)add(a,b);
            else add(b,a);
        }
    }
    cout<<"Yes"<<endl;
    rep(i,ans.size())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 ll;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using P=pair<int,int>;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,d[3];cin>>n;
    rep(i,3)cin>>d[i];
    vector<P> q;
    rep(i,n){
        string ss;cin>>ss;
        int a,b;
        if(ss=="AB")a=0,b=1;
        else if(ss=="BC")a=1,b=2;
        else a=0,b=2;
        q.push_back({a,b});
    }
    q.push_back({0,1});
    string ans;
    auto add=[&](int a,int b){
        d[a]++;d[b]--;
        ans+='A'+a;
    };
    rep(i,n){
        int a=q[i].first;
        int b=q[i].second;
        if(!d[a]&&!d[b]){
            cout<<"No"<<endl;
            return 0;
        }
        else if(!d[a])add(a,b);
        else if(!d[b])add(b,a);
        else {
            int na=q[i+1].first;
            int nb=q[i+1].second;
            if(a==na||a==nb)add(a,b);
            else add(b,a);
        }
    }
    cout<<"Yes"<<endl;
    rep(i,ans.size())cout<<ans[i]<<endl;
    return 0;
}

```