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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+10;
const ll inf = 1e9+7;
string s;
pair<int,int>p[maxn];
vector<char>ans;
int main() {
    ios::sync_with_stdio(0);
    int n,a[5];
    cin>>n>>a[0]>>a[1]>>a[2];
    for(int i=1; i<=n; i++) {
        cin>>s;
        p[i]= {s[0]-'A',s[1]-'A'};
    }
    for(int i=1; i<=n; i++) {
        int x = p[i].first;
        int y = p[i].second;
        if(a[x]==0&&a[y]==0) {
            cout<<"No"<<endl;
            return 0;
        }
        int d = x;
        if(a[x]==1&&a[y]==1) {
            if(i<n&&p[i+1].first==y||p[i+1].second==y)
                d =y;
        } else if(a[x]>=a[y])
            d=y;
        ans.push_back(d+'A');
        if(d==x)
            a[x]++,a[y]--;
        else
            a[y]++,a[x]--;
    }
    cout<<"Yes"<<endl;
    for(auto j:ans)
        cout<<j<<endl;
    return 0;
}