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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int main ()
{
	int a, b;
	cin >> a >> b;
	if(a == 0 || b == 0){
		cout << 0;
	}
	else if(a == 1 && b >= 1){
		cout << 1;
	}
	else if( a == 2 && b >= 2){
		cout << 2;
	}
	else if( a == 3 && b >= 3){
		cout << 3;
	}
else if( a == 4 && b >= 4){
		cout << 4;
	}
else if( a == 5 && b >= 5){
		cout << 5;
	}
else if( a == 6 && b >= 6){
		cout << 6;
	}
else if( a == 7 && b >= 7){
		cout << 7;
	}
	else if( a == 8 && b >= 8){
		cout << 8;
	}
else if( a == 9 && b >= 9){
		cout << 9;
	}
else if( a == 10 && b >= 10){
		cout << 10;
	}
else if( a == 11 && b >= 11){
		cout << 11;
	}
else if( a == 12 && b >= 12){
		cout << 12;
	}
	

		return 0;
}