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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define INF 1e12

#define PB push_back
#define PF push_front
#define fi first
#define se second
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vpi vector<pii>
#define vll vector<ll>
#define vpl vector<pll>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define MX(x) *max_element(all(x))
#define MN(x) *min_element(all(x))
#define ios ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define pr_d(x) cout << fixed << setprecision(15) << x << endl
#define ud(c, x) distance(c.begin(), upper_bound(all(c), x))
#define ld(c, x) distance(c.begin(), lower_bound(all(c), x))
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define rep3(i, n) for (int i = (n - 1); i >= 0; --i)
#define rep4(i, a, b) for (int i = (a); i > (b); --i)
#define pb push_back
#define out(x) cout << x << "\n"
bool odd(int i) { return i % 2; }

#define all(v) v.begin(), v.end()
#define size(x) int(x.size())

int gcd(int a, int b) { return __gcd(a, b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
void Yes_No(bool f) {
  if (f)
    printf("Yes\n");
  else
    printf("No\n");
}
void YES_NO(bool f) {
  if (f)
    printf("YES\n");
  else
    printf("NO\n");
}

template <typename T>
void deb1(T x) {
  cout << x << "\n";
}

template <typename T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
//-------------------ここから回答する-----------------------

void solve(void) {
  int n;
  cin >> n;
  vi v(n);
  rep(i, n) cin >> v[i];
  sort(all(v));
  out(*v.end() - *v.begin());
}
int main(void) { solve(); }