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 A {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int m = scan.nextInt();
		int d = scan.nextInt();
		scan.close();
		int ans=0;
		switch (m) {
		
			case 1:
			case 3:
			case 5:
			case 7:
			case 8:
			case 10:
			case 12:
			m=31;
			break;
			case 2:
			m=28;
			break;
			case 4:
			case 6:
			case 9:
			case 11:
			m=30;
			break;

	}
	switch (d){
	case 1:
		if(m%7==0){
			ans=m/7;
		}else{
			ans=m/7+1;
		}
		break;
	case 2:
		if((m-6)%7==0){
			ans=(m-6)/7+1;
		}else{
			ans=(m-6)/7+2;
		}
		break;
	case 3:
		if((m-5)%7==0){
			ans=(m-5)/7+1;
		}else{
			ans=(m-5)/7+2;
		}
		break;
	case 4:
		if((m-4)%7==0){
			ans=(m-4)/7+1;
		}else{
			ans=(m-4)/7+2;
		}
		break;
	case 5:
		if((m-3)%7==0){
			ans=(m-3)/7+1;
		}else{
			ans=(m-3)/7+2;
		}
		break;
	case 6:
		if((m-2)%7==0){
			ans=(m-2)/7+1;
		}else{
			ans=(m-2)/7+2;
		}
		break;
	case 7:
		if((m-1)%7==0){
			ans=(m-1)/7+1;
		}else{
			ans=(m-1)/7+2;
		}
		break;
	
	
	}
		
		System.out.println(ans);

		
	}

}
