#include <cstdio>
#include <algorithm>
#define N 100005

int n;
int c[5];
char s[N][5], ans[N];

int main() {
    int *a = c - 'A';
    scanf("%d%d%d%d", &n, a + 'A', a + 'B', a + 'C');
    for (int i = 1; i <= n; ++i) scanf("%s", s[i]);
    for (int i = 1; i <= n; ++i)
        if (!a[s[i][0]] && !a[s[i][1]]) {
            puts("No");
            return 0;
        } else if (a[s[i][0]] < a[s[i][1]] || (a[s[i][0]] == a[s[i][1]] && i < n && (s[i][0] == s[i + 1][0] && s[i][1] != s[i + 1][1] || s[i][0] == s[i + 1][1] && s[i][1] != s[i + 1][0]))) ++a[s[i][0]], --a[s[i][1]], ans[i] = s[i][0];
        else --a[s[i][0]], ++a[s[i][1]], ans[i] = s[i][1];
    puts("Yes");
    for (int i = 1; i <= n; ++i) printf("%c\n", ans[i]);
}

### 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 <cstdio>
#include <algorithm>
#define N 100005

int n;
int c[5];
char s[N][5], ans[N];

int main() {
    int *a = c - 'A';
    scanf("%d%d%d%d", &n, a + 'A', a + 'B', a + 'C');
    for (int i = 1; i <= n; ++i) scanf("%s", s[i]);
    for (int i = 1; i <= n; ++i)
        if (!a[s[i][0]] && !a[s[i][1]]) {
            puts("No");
            return 0;
        } else if (a[s[i][0]] < a[s[i][1]] || (a[s[i][0]] == a[s[i][1]] && i < n && (s[i][0] == s[i + 1][0] && s[i][1] != s[i + 1][1] || s[i][0] == s[i + 1][1] && s[i][1] != s[i + 1][0]))) ++a[s[i][0]], --a[s[i][1]], ans[i] = s[i][0];
        else --a[s[i][0]], ++a[s[i][1]], ans[i] = s[i][1];
    puts("Yes");
    for (int i = 1; i <= n; ++i) printf("%c\n", ans[i]);
}

```