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.
#include <bits/stdc++.h>
using namespace std;
const int xx[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int yy[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int kx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int ky[] = {1, 2, 2, 1, -1, -2, -2, -1};
int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int cen[10][302];
int main() {
  int i, j, k, n, d;
  scanf("%d%d", &n, &d);
  int x = d;
  k = 1;
  for (i = 1; k <= month[n]; i++) {
    for (j = x; j <= 7 && k <= month[n]; j++) cen[j][i] = k++;
    x = 1;
  }
  printf("%d\n", i - 1);
  return 0;
}
