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: Accepted)

#include<bits/stdc++.h>
//#include<atcoder/all>
using namespace std;
//using namespace atcoder;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define all(x) (x).begin(),(x).end()
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[8]={1,0,-1,0,1,1,-1,-1};
const ll dx[8]={0,-1,0,1,1,-1,1,-1};
using Graph = vector<vector<int>>;

// dijkstra　
struct edge{ll to, cost;};
typedef pair<ll,ll> P;
struct graph{
  ll V;
  vector<vector<edge> > G;
  vector<ll> d;

  graph(ll n){
    init(n);
  }

  void init(ll n){
    V = n;
    G.resize(V);
    d.resize(V);
    rep(i,V){
      d[i] = INF;
    }
  }

  void add_edge(ll s, ll t, ll cost){
    edge e;
    e.to = t, e.cost = cost;
    G[s].push_back(e);
  }

  void dijkstra(ll s){
    rep(i,V){
      d[i] = INF;
    }
    d[s] = 0;
    priority_queue<P,vector<P>, greater<P> > que;
    que.push(P(0,s));
    while(!que.empty()){
      P p = que.top(); que.pop();
      ll v = p.second;
      if(d[v]<p.first) continue;
      for(auto e : G[v]){
        if(d[e.to]>d[v]+e.cost){
          d[e.to] = d[v]+e.cost;
          que.push(P(d[e.to],e.to));
        }
      }
    }
  }
};
// dijkstra end


class UnionFind
{
public:
  ll par[100005];
  ll depth[100005];
  ll nGroup[100005];
 
  UnionFind(ll n) {
    init(n);
  }
 
  void init(ll n) {
    for(ll i=0; i<n; i++) {
      par[i] = i;
      depth[i] = 0;
      nGroup[i] = 1;
    }
  }
 
  ll root(ll x) {
    if(par[x] == x) {
      return x;
    } else {
      return par[x] = root(par[x]);
    }
  }
 
  bool same(ll x, ll y) {
    return root(x) == root(y);
  }
 
  void unite(ll x, ll y) {
    x = root(x);
    y = root(y);
    if(x == y) return;
 
    if(depth[x] < depth[y]) {
      par[x] = y;
      nGroup[y] += nGroup[x];
      nGroup[x] = 0;
    } else {
      par[y] = x;
      nGroup[x] += nGroup[y];
      nGroup[y] = 0;
      if(depth[x] == depth[y])
        depth[x]++;
    }
  }
};
// unionfind end

// nCr
const ll MAX = 500010;
ll fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// nCr end

// tree DP
vector<ll> depth;
vector<ll> f;
vector<ll> g;
void dfs(const Graph &G, ll v, ll p, ll d) {
    depth[v] = d;
    for (auto nv : G[v]) {
        if (nv == p) continue; 
        dfs(G, nv, v, d+1);
    }


    f[v] = 1;
    g[v] = 1; 
    for (auto c : G[v]) {
        if (c == p) continue;
        f[v]*=g[c];
        f[v]%=MOD;
        g[v]*=f[c];
        g[v]%=MOD;
        }
    f[v]+=g[v];
    f[v]%=MOD;
}
// tree DP end


template<ll MOD> struct Fp {
    ll val;
    constexpr Fp(ll v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr ll getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        ll a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            ll t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, ll n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
using mint = Fp<MOD>;

mint calc(ll N, ll K) {
    mint res = 1;
    for (ll n = 0; n < K; ++n) {
        res *= (N - n);
        res /= (n + 1);
    }
    return res;
}
mint COM(ll n, ll k){
    if (k == 0) return 1;
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
} 


int main() {
  ll n; cin>>n;
  ll a[n]={};
  rep(i,n) cin>>a[i];
  sort(a,a+n);
  cout << a[n-1] - a[0] << endl;
}