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>
const int INF = 1e9;
const int MOD = 1e9+7;
using LL = long long;
const LL LINF = 1e18;
using namespace std;
#define COUT(v) cout<<(v)<<endl
#define CIN(n)  int(n);cin >> (n)
#define SCIN(n) string(n);cin >> (n)
#define YES(n) cout<<((n)? "YES" : "NO")<<endl
#define Yes(n) cout<<((n)? "Yes" : "No")<<endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) <<endl

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)

#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(obj) (obj).begin(),(obj).end()

#define P pair<int,int>
#define I vector<int>
#define S set<int>
#define pb(v) push_back(v)
#define V vector
#define rt return
#define rmsame(a) sort(ALL(a)),a.erase(unique(ALL(a)), a.end())
class Edge{
public:
    int from,to,value;
    Edge(int a,int b,int c){
        from = a;
        to = b;
        value = c;
    }
    Edge(int a,int b){
        from = a;
        to = b;
    }
};
int main(){
    CIN(N);CIN(B);
    COUT(N-1+(N<=B?+1:+0));
    return 0;
}