Problem:
Score : 200 points

Problem Statement

It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts.

There are N houses along TopCoDeer street. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses.

Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.

Constraints

1 ≤ N ≤ 100

0 ≤ a_i ≤ 1000

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum distance to be traveled.

Sample Input 1

4
2 3 7 9

Sample Output 1

7

The travel distance of 7 can be achieved by starting at coordinate 9 and traveling straight to coordinate 2.

It is not possible to do with a travel distance of less than 7, and thus 7 is the minimum distance to be traveled.

Sample Input 2

8
3 1 4 1 5 9 2 6

Sample Output 2

8

There may be more than one house at a position.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using static System.Console;
using static System.Math;

namespace At
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = ReadInt();
            var a = ReadInts();

            WriteLine(a.Max() - a.Min());
        }

        private static string Read() { return ReadLine(); }
        private static string[] Reads() { return (Read().Split()); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static double ReadDouble() { return double.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
        private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }

        //最大公約数
        
        public static int Gcd(int a, int b)
        {
            if (a < b)
            {
                return Gcd(b, a);
            }
            while (b != 0)
            {
                var reminder = a % b;
                a = b;
                b = reminder;
            }
            return a;
        }

        //最小公倍数
        public static int Lcm(int a, int b)
        {
            return a * b / Gcd(a, b);
        }

    }
}