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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define Abs(x)((x) < 0 ? (x) * -1 : (x))

#define rep(x, y) for ((x) = 0; (x) < (y); (x)++)
#define repin(x, y) for ((x) = 0; (x) <= (y); (x)++)
#define nep(x, y) for ((x) = (y) - 1; 0 <= (x); (x)--)
#define nepi(x, y, z) for ((x) = (y) - 1; (z) <= (x); (x)--)
#define repi(x, y, z) for ((x) = (z); (x) < (y); (x)++)
#define repiin(x, y, z) for ((x) = (z); (x) <= (y); (x)++)
#define reps(x, y, z) for ((x) = 0; (x) < (y); (x) += (z))
#define repis(x, y, z, s) for ((x) = (z); (x) < (y); (x) += (s))
#define repiins(x, y, z, s) for ((x) = (z); (x) <= (y); (x) += (s))
#define repit(x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)
#define nepit(x) for (__typeof((x).rbegin()) it = (x).rbegin(); it != (x).rend(); it++)

#define All(x) (x).begin(),(x).end()

#define Mem0(x) memset(x, 0, sizeof(x))
#define Mem1(x) memset(x, -1, sizeof(x))

#define Unique(v) v.resize(unique(All(v)) - v.begin())

#define peq(p0, p1)(p0.first == p1.first && p0.second == p1.second)

#define End '\n'
#define Out(x) cout<<(x)<<End
#define OutP(p, sep, end) cout<<p.first<<sep<<p.second<<end

template<typename T> inline bool Max(T &x, const T &y) { return x < y ? x = y, 1 : 0; }
template<typename T> inline bool Min(T &x, const T &y) { return x > y ? x = y, 1 : 0; }
template<typename T> using V = vector<T>;
template<typename T> using VV = V<V<T> >;

#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define lb std::lower_bound
#define ub std::upper_bound

#define lbi(v, x) lb(All(v), (x))-v.begin()
#define ubi(v, x) ub(All(v), (x))-v.begin()

static const ll MOD = 1e9 + 7;
static const double PI = 3.141592653589793;

#define LOCAL 0

int main()
{
#if LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("debug.txt", "w", stderr);
#endif

    cin.tie(0);
    ios::sync_with_stdio(false);
	//std::cout.precision(18);

    int n;

	cin >> n;

    int a[n];

	int i;

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

	sort(a, a + n);

	Out(a[n - 1] - a[0]);
	
	return 0;
}