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: Accepted)

#include <stdio.h>

int days(int month) {
  int d;
  switch (month) {
    case 1: case 3: case 5: case 7: case 8: case 10: case 12:
      d = 31;
      break;
    case 4: case 6: case 9: case 11:
      d = 30;
      break;
    case 2:
      d = 29;
      break;
  }
  return d;
}

int main(void) {
  int i, j, a, b, c = 0;
  
  scanf("%d %d", &a, &b);
  
  for (i = 1; i <= a; i++)
    for (j = 1; i == a ? j <= b : j <= days(i); j++)
      if (i == j)
        c++;
  
  printf("%d", c);
  
  return 0;
}