#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, d;
  scanf("%d %d", &m, &d);
  if (m == 2) {
    if (d == 1)
      printf("4\n");
    else
      printf("5\n");
  } else if (m == 4 || m == 6 || m == 9 || m == 11) {
    if (d == 7)
      printf("6\n");
    else
      printf("5\n");
  } else {
    if (d == 6 || d == 7)
      printf("6\n");
    else
      printf("5\n");
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:

<image>

Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.

Input

The only line contain two integers m and d (1 ≤ m ≤ 12, 1 ≤ d ≤ 7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).

Output

Print single integer: the number of columns the table should have.

Examples

Input

1 7


Output

6


Input

1 1


Output

5


Input

11 6


Output

5

Note

The first example corresponds to the January 2017 shown on the picture in the statements.

In the second example 1-st January is Monday, so the whole month fits into 5 columns.

In the third example 1-st November is Saturday and 5 columns is enough.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, d;
  scanf("%d %d", &m, &d);
  if (m == 2) {
    if (d == 1)
      printf("4\n");
    else
      printf("5\n");
  } else if (m == 4 || m == 6 || m == 9 || m == 11) {
    if (d == 7)
      printf("6\n");
    else
      printf("5\n");
  } else {
    if (d == 6 || d == 7)
      printf("6\n");
    else
      printf("5\n");
  }
  return 0;
}

```