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)

#pragma region header
#define _GLIBCXX_DEBUG 
#include <bitset>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <cctype>
#include <assert.h>
#include <stdlib.h>
#include <stdio.h>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <limits>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <math.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define ts to_string
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vs vector <string>
#define vvs vector <vector<string>>
#define pq  priority_queue<int>
#define smallpq  priority_queue<int, vector<int>, greater<int>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
constexpr ll mod = 1e9 + 7;
constexpr ll inf = INT32_MAX / 2;
constexpr ll INF = LLONG_MAX / 2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */
/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if (b == 0) return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a * b / gcd(a, b);
}
bool isprime(ll a)//素数判定、primeならtrue,違うならfalse
{
    if (a < 2) return false;
    else if (a == 2) return true;
    else if (a % 2 == 0) return false;
    double m = sqrt(a);
    for (int i = 3; i <= m; i += 2)
    {
        if (a % i == 0)
        {
            return false;
        }
    }
    // 素数である
    return true;
}
ll keta(ll n) {        //桁数を求める
    if (n == 0) return 1;
    int count = 0;
    while (n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
/* ----------------------end----------------------- */
#pragma endregion
signed main() {
    int A, B;
    cin >> A >> B;
    if (A > B)cout << A - 1 << endl;
    else cout << A << endl;
}