Problem:
Score: 100 points

Problem Statement

In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.

For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.

In this country, a date is called Takahashi when the month and the day are equal as numbers. For example, 5-5 is Takahashi.

How many days from 2018-1-1 through 2018-a-b are Takahashi?

Constraints

a is an integer between 1 and 12 (inclusive).

b is an integer between 1 and 31 (inclusive).

2018-a-b is a valid date in Gregorian calendar.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the number of days from 2018-1-1 through 2018-a-b that are Takahashi.

Sample Input 1

5 5

Sample Output 1

5

There are five days that are Takahashi: 1-1, 2-2, 3-3, 4-4 and 5-5.

Sample Input 2

2 1

Sample Output 2

1

There is only one day that is Takahashi: 1-1.

Sample Input 3

11 30

Sample Output 3

11

There are eleven days that are Takahashi: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9, 10-10 and 11-11.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
 int p(int n)
 {
 	int i, br=0;
	 for(i=2;i<n;i++)
	 {
	 	if (n%i==0)
br=br+1;
	 }
 if(br==0)
 return 1;
 else
 return 0;
 }
 main()
 {
 	int n, s=0, br=0, i;
 	scanf("%d", &n);
 	for(i=2;i<55555;i++)
	 {
	 	if (br==n)
	 	break;
		 if (p(i)==1)
	 	{
	 		if(br==5)
	 		{
	 			s=s+i;
	 			if(p(s)==0)
	 			{
	
		br=br+1;		 
	 			printf("%d ", i);
		}
			 else
			 s=s-i;
			 }
		 else if(br<5)
		 {
		 	printf("%d ", i);
		 	br=br+1;
		 s=s+i;
		 }
		else if (br>5)
		 {
		 	printf("%d ", i);
		 	br=br+1;
		 }
		 }
	 }
 	 return 0;
 }