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)

/*
ID: anonymo14
TASK: wormhole
LANG: C++                 
*/
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int,int>> vpii;

#define F first
#define S second
#define PU push
#define PUF push_front
#define PUB push_back
#define PO pop
#define POF pop_front
#define POB pop_back

#define REP(i,a,b) for(int i=a; i<=b; i++)

#define MOD 1000000007

void solve(int test_case) {//222
	int a,b;
	cin>>a>>b;
	if(b<a)cout<<a-1;
	else cout<<a;
}

int main() {
	
	////// FILE BASED IO////
	//freopen("wormhole.in", "r", stdin);
	//freopen("wormhole.out", "w", stdout);
	///////////////
	
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t=1;
	//cin>>t;
	REP(i,1,t) {
		solve(i);
	}
	return 0;
}