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;
long long m, d;
long long a[1000][1000];
long long me[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m >> d;
  long long cnt = 1, cnt1 = 0;
  long long pr = 0;
  long long ok = 0;
  for (long long i = 1; i <= 1000; i++) {
    if (ok == 0) {
      for (long long j = d; j <= 7; j++) {
        a[i][j] = cnt;
        cnt++;
        if (cnt - 1 == me[m - 1]) {
          cout << i;
          return 0;
        }
      }
      ok = 1;
    } else {
      for (long long j = 1; j <= 7; j++) {
        a[i][j] = cnt;
        cnt++;
        if (cnt - 1 == me[m - 1]) {
          cout << i;
          return 0;
        }
      }
    }
  }
}
