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>
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
// #include <numeric>
#include <complex>

#define FAST ios_base::sync_with_stdio(false); cin.tie(0)
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define repr(i, a, b) for(int i=int(a); i>=int(b); i--)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
typedef long long int ll;
#define iceil(x,y) (x+y-1)/y
#define min(x,y)(x<y?x:y)

using namespace std;
typedef pair<int, int> P;

int main() {
	FAST;

	int a, b;
	cin >> a >> b;

	int res;
	if (b>12) res = a;
	else res = (a<b? a:b);
	cout << res;

	return 0;
}