#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[3],n;
char s[N][3];
vector<char>ans;
int dfs(int now){
    if(a[0] < 0 || a[1] < 0 || a[2] < 0)
        return false;
    if(now == n + 1)
        return true;

    int x = s[now][0] - 'A';
    int y = s[now][1] - 'A';
    if(a[x] > a[y])
        swap(x,y);
    a[x]++,a[y]--;
    ans.push_back(x+'A');
    if(dfs(now+1)) return true;
    ans.pop_back();
    ans.push_back(y+'A');
    a[x]-=2,a[y]+=2;
    if(dfs(now+1)) return true;
    ans.pop_back();
    a[x]++,a[y]--;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i=0;i<3;i++){
        cin >> a[i];
    }
    for(int i=1;i<=n;i++){
        cin >> s[i];
    }
    if(dfs(1)){
        cout << "Yes\n";
        for(auto i:ans)
            cout << i << '\n';
        return 0;
    }
    cout << "No\n";
    return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e5 + 10;
int a[3],n;
char s[N][3];
vector<char>ans;
int dfs(int now){
    if(a[0] < 0 || a[1] < 0 || a[2] < 0)
        return false;
    if(now == n + 1)
        return true;

    int x = s[now][0] - 'A';
    int y = s[now][1] - 'A';
    if(a[x] > a[y])
        swap(x,y);
    a[x]++,a[y]--;
    ans.push_back(x+'A');
    if(dfs(now+1)) return true;
    ans.pop_back();
    ans.push_back(y+'A');
    a[x]-=2,a[y]+=2;
    if(dfs(now+1)) return true;
    ans.pop_back();
    a[x]++,a[y]--;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i=0;i<3;i++){
        cin >> a[i];
    }
    for(int i=1;i<=n;i++){
        cin >> s[i];
    }
    if(dfs(1)){
        cout << "Yes\n";
        for(auto i:ans)
            cout << i << '\n';
        return 0;
    }
    cout << "No\n";
    return 0;
}

```