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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define ct(n) cout << n << endl;
#define cts(n) cout << n << " ";
#define cte cout << endl;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define repo(i,o,n) for(int i=o;i<n;i++)
#define repm(i,n) for(int i=n-1;i>=0;i--)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end());
#define vdelete(v) (v).erase(unique(all(v)), (v).end());
#define vcopy(v1,v2) copy((v1).begin(), (v1).end(), back_inserter((v2)));
#define pb(n) push_back(n);

int a,b;

main() {
    cin >> a >> b;
    if (a>b) a--;
    ct(a)
}