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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>

#define rep(i, n) for(int i=0; i<(n); ++i)
#define max  200001

//qsort(str, n, sizeof(int), up_comp);
int up_comp(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_comp(const void *a, const void *b){return *(int*)b - *(int*)a;}
int int_sort( const void * a , const void * b ) {return strcmp(( char * )a , ( char * )b );}
int count_digit(int a){int i=0;while(a!=0){a=a/10;i++;}return i;}
int euclid( int a, int b ){int temp;if(a<b){temp=a;a=b;b=temp;}if(b<1)return -1;if(a%b==0) return b;return euclid(b,a%b);}

int main(void){
 
    int a, b;
    int ans;
    
    scanf("%d%d", &a, &b);
    
    if (b>=a) {
        ans=a;
    } else {
        ans=a-1;
    }
    
    printf("%d", ans);
    
    return 0;
}