#include <bits/stdc++.h>
using namespace std;
void pause() {}
int dv(int a, int b) {
  double s;
  s = (a - (a % b)) / b;
  int d;
  d = (int)(s + 0.5);
  return d;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  int k, i;
  k = 1;
  if (m == 1) k = 0;
  for (i = 1; i <= 31; i++) {
    if (m == 1) k++;
    m++;
    if (m == 8) m = 1;
    if (n == 4 || n == 6 || n == 9 || n == 11) {
      if (i == 30) break;
    }
    if (n == 2 && i == 28) break;
  }
  cout << k;
  pause();
  return (0);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
void pause() {}
int dv(int a, int b) {
  double s;
  s = (a - (a % b)) / b;
  int d;
  d = (int)(s + 0.5);
  return d;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  int k, i;
  k = 1;
  if (m == 1) k = 0;
  for (i = 1; i <= 31; i++) {
    if (m == 1) k++;
    m++;
    if (m == 8) m = 1;
    if (n == 4 || n == 6 || n == 9 || n == 11) {
      if (i == 30) break;
    }
    if (n == 2 && i == 28) break;
  }
  cout << k;
  pause();
  return (0);
}

```