Problem:
Score : 200 points

Problem Statement

You are given an integer sequence A of length N.
Find the maximum absolute difference of two elements (with different indices) in A.

Constraints

2 \leq N \leq 100

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum absolute difference of two elements (with different indices) in A.

Sample Input 1

4
1 4 6 3

Sample Output 1

5

The maximum absolute difference of two elements is A_3-A_1=6-1=5.

Sample Input 2

2
1000000000 1

Sample Output 2

999999999

Sample Input 3

5
1 1 1 1 1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

// include
// ------------------------------------------------
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

// func
// ------------------------------------------------
int CalcSumOfDigit(int n);      // 各桁の和を計算する。
string upper(string str);       // 英字を大文字に変換する。
string lower(string str);       // 英字を小文字に変換する。

// define
// ------------------------------------------------
#define all(a)        (a).begin(),(a).end()
#define rall(a)       (a).rbegin(),(a).rend()
#define sz(a)         int((a).size())
#define rep(i,n)      for(int(i)=0;(i)<(n);(i)++)
#define repe(i,n)     for(int(i)=0;(i)<=(n);(i)++)
#define vsort(v)      sort((v).begin(),(v).end())
#define rvsort(v)     sort(rall((v)))
#define vi            vector<int>
#define GCD(a,b)      __gcd((a),(b))
#define LCM(a,b)      (a)/GCD((a),(b))*(b)
#define kiriage(a,b)  ((a)+(b)-1)/(b)

const int INF = 1e9;

// code
// ------------------------------------------------
int main() {

  int n;
  vi a(n);
  rep(i,n) cin >> a[i];

  vsort(a);

  cout << a[n-1] - a[0] << endl;
  return 0;
}
// funcの実体
// ------------------------------------------------
int CalcSumOfDigit(int n)
{
  int s = 0;
  while(n)
  {
    s += n % 10;
    n = n / 10;
  }
  return s;
}

string upper(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(97 <= *itr && *itr <= 122)
    {
      *itr = *itr - 32;
    }
  }
  return str;
}

string lower(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(65 <= *itr && *itr <= 90)
    {
      *itr = *itr + 32;
    }
  }
  return str;
}