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 main(void){
  int a,b,ans;
  scanf("%d %d",&a,&b);
  if (a > 11 && b > 11) {
    ans = 12;
  } else if (a > 11 || (a > 10 && b > 10)) {
    ans = 11;
  } else if (a > 10 || (a > 9 && b > 9)) {
    ans = 10;
  } else if (a > 9 || (a > 8 && b > 8)) {
    ans = 9;
  } else if (a > 8 || (a > 7 && b > 7)) {
    ans = 8;
  } else if (a > 7 || (a > 6 && b > 6)) {
    ans = 7;
  } else if (a > 6 || (a > 5 && b > 5)) {
    ans = 6;
  } else if (a > 5 || (a > 4 && b > 4)) {
    ans = 5;
  } else if (a > 4 || (a > 3 && b > 3)) {
    ans = 4;
  } else if (a > 3 || (a > 2 && b > 2)) {
    ans = 3;
  } else if (a > 2 || (a > 1 && b > 1)) {
    ans = 2;
  } else {
    ans = 1;
  }
  printf("%d\n",ans);
  return 0;
}