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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Text.RegularExpressions;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;

namespace ConsoleApp2
{
    class Program
    {
        class ReverseComparer : IComparer<int>
        {
            public int Compare(int x, int y)
            {
                // 既定のComparerとは反対の結果を返す
                return Comparer<int>.Default.Compare(y, x);
            }
        }

        static void Main(string[] args)
        {

            List<int> values = new List<int>();
            string str = Console.ReadLine(); //標準入力
            //string str = "11 11";
            string[] str2 = str.Split(' ');
            int a = int.Parse(str2[0]);
            int b = int.Parse(str2[1]);

            char[,] masu = new char[a + 2, b + 2];

            for (int i = 0; i < b; i++)
            {
                str = Console.ReadLine();
                char[] c = str.ToCharArray();
                for (int i2 = 0; i2 < c.Length; i2++)
                {
                    masu[i + 1, i2 + 1] = c[i2];
                }
            }





            for (int y = 1; y < b + 1; y++)
            {
                for (int x = 1; x < a + 1; x++)
                {
                    if (masu[y, x] == '#')
                    {
                        if (masu[y - 1, x] != '#' && masu[y + 1, x] != '#' && masu[y, x - 1] != '#' && masu[y, x + 1] != '#')
                        {
                            Console.WriteLine("No");
                            return;
                        }
                    }
                }
            }

            Console.WriteLine("Yes");
        }
    }
}