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.
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int m = scan.nextInt();
		int d = scan.nextInt();
		d--;
		int ans=0;
		/*
		1	31
		2	59
		3	90
		4	120
		5	151
		6	181
		7	212
		8	243
		9	273
		10	304
		11	334
		12	365

		*/
		
		if ((m==1)||(m==3)||(m==5)||(m==7)||(m==8)||(m==10)||(m==12)){
			if ((31+d)%7 == 0){ans = ((31+d)/7);}
			else {ans = ((31+d)/7)+1;}
		}
		else if (m==2){
			if ((28+d)%7 == 0){ans = ((28+d)/7);}
			else {ans = ((28+d)/7)+1;}
		}
		else {
			if ((30+d)%7 == 0){ans = ((30+d)/7);}
			else {ans = ((30+d)/7)+1;}
		}
		
		
		System.out.println(ans);
	}

}

			 	 	 		   	 	 				 									