#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int sz = 1e5+5;

string ans, s[sz];

int main()
{
    int n, a, b, c;
    cin >> n >> a >> b >> c;

    bool yes = 1;
    for(int i=0; i<n; i++) cin >> s[i];
    for(int i=0; i<n; i++) {
        if(s[i] == "AB") {
            if(a < b) ans += 'A', a++, b--;
            else if(a > b) ans += 'B', b++, a--;
            else if(!c && s[i+1] == "AC") ans += 'A', a++, b--;
            else ans += 'B', b++, a--;
        }
        else if(s[i] == "BC") {
            if(b < c) ans += 'B', b++, c--;
            else if(b > c) ans += 'C', c++, b--;
            else if(!a && s[i+1] == "AB") ans += 'B', b++, c--;
            else ans += 'C', c++, b--;
        }
        else {
            if(c < a) ans += 'C', c++, a--;
            else if(c > a) ans += 'A', a++, c--;
            else if(!b && s[i+1] == "BC") ans += 'C', c++, a--;
            else ans += 'A', a++, c--;
        }
        if(a < 0 || b < 0 || c < 0) yes = 0;
    }

    if(yes) {
        puts("Yes");
        for(char x : ans) printf("%c\n", x);
    }
    else puts("No");
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
typedef long long ll;
const int sz = 1e5+5;

string ans, s[sz];

int main()
{
    int n, a, b, c;
    cin >> n >> a >> b >> c;

    bool yes = 1;
    for(int i=0; i<n; i++) cin >> s[i];
    for(int i=0; i<n; i++) {
        if(s[i] == "AB") {
            if(a < b) ans += 'A', a++, b--;
            else if(a > b) ans += 'B', b++, a--;
            else if(!c && s[i+1] == "AC") ans += 'A', a++, b--;
            else ans += 'B', b++, a--;
        }
        else if(s[i] == "BC") {
            if(b < c) ans += 'B', b++, c--;
            else if(b > c) ans += 'C', c++, b--;
            else if(!a && s[i+1] == "AB") ans += 'B', b++, c--;
            else ans += 'C', c++, b--;
        }
        else {
            if(c < a) ans += 'C', c++, a--;
            else if(c > a) ans += 'A', a++, c--;
            else if(!b && s[i+1] == "BC") ans += 'C', c++, a--;
            else ans += 'A', a++, c--;
        }
        if(a < 0 || b < 0 || c < 0) yes = 0;
    }

    if(yes) {
        puts("Yes");
        for(char x : ans) printf("%c\n", x);
    }
    else puts("No");
}

```