Problem: [Æsir - CHAOS](https://soundcloud.com/kivawu/aesir-chaos)

[Æsir - V.](https://soundcloud.com/kivawu/aesir-v)

"Everything has been planned out. No more hidden concerns. The condition of Cytus is also perfect.

The time right now...... 00:01:12......

It's time."

The emotion samples are now sufficient. After almost 3 years, it's time for Ivy to awake her bonded sister, Vanessa.

The system inside A.R.C.'s Library core can be considered as an undirected graph with infinite number of processing nodes, numbered with all positive integers (1, 2, 3, …). The node with a number x (x > 1), is directly connected with a node with number (x)/(f(x)), with f(x) being the lowest prime divisor of x.

Vanessa's mind is divided into n fragments. Due to more than 500 years of coma, the fragments have been scattered: the i-th fragment is now located at the node with a number k_i! (a factorial of k_i).

To maximize the chance of successful awakening, Ivy decides to place the samples in a node P, so that the total length of paths from each fragment to P is smallest possible. If there are multiple fragments located at the same node, the path from that node to P needs to be counted multiple times.

In the world of zeros and ones, such a requirement is very simple for Ivy. Not longer than a second later, she has already figured out such a node.

But for a mere human like you, is this still possible?

For simplicity, please answer the minimal sum of paths' lengths from every fragment to the emotion samples' assembly node P.

Input

The first line contains an integer n (1 ≤ n ≤ 10^6) — number of fragments of Vanessa's mind.

The second line contains n integers: k_1, k_2, …, k_n (0 ≤ k_i ≤ 5000), denoting the nodes where fragments of Vanessa's mind are located: the i-th fragment is at the node with a number k_i!.

Output

Print a single integer, denoting the minimal sum of path from every fragment to the node with the emotion samples (a.k.a. node P).

As a reminder, if there are multiple fragments at the same node, the distance from that node to P needs to be counted multiple times as well.

Examples

Input


3
2 1 4


Output


5


Input


4
3 1 4 4


Output


6


Input


4
3 1 4 1


Output


6


Input


5
3 1 4 1 5


Output


11

Note

Considering the first 24 nodes of the system, the node network will look as follows (the nodes 1!, 2!, 3!, 4! are drawn bold):

<image>

For the first example, Ivy will place the emotion samples at the node 1. From here:

  * The distance from Vanessa's first fragment to the node 1 is 1. 
  * The distance from Vanessa's second fragment to the node 1 is 0. 
  * The distance from Vanessa's third fragment to the node 1 is 4. 



The total length is 5.

For the second example, the assembly node will be 6. From here:

  * The distance from Vanessa's first fragment to the node 6 is 0. 
  * The distance from Vanessa's second fragment to the node 6 is 2. 
  * The distance from Vanessa's third fragment to the node 6 is 2. 
  * The distance from Vanessa's fourth fragment to the node 6 is again 2. 



The total path length is 6.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template <class t>
using vvc = vc<vc<t>>;
using pi = pair<ll, ll>;
using vi = vc<ll>;
template <class t, class u>
ostream& operator<<(ostream& os, const pair<t, u>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class t>
ostream& operator<<(ostream& os, const vc<t>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
using uint = unsigned;
using ull = unsigned long long;
template <class t, size_t n>
ostream& operator<<(ostream& os, const array<t, n>& first) {
  return os << vc<t>(first.begin(), first.end());
}
template <ll i, class T>
void print_tuple(ostream&, const T&) {}
template <ll i, class T, class H, class... Args>
void print_tuple(ostream& os, const T& t) {
  if (i) os << ",";
  os << get<i>(t);
  print_tuple<i + 1, T, Args...>(os, t);
}
template <class... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
  os << "{";
  print_tuple<0, tuple<Args...>, Args...>(os, t);
  return os << "}";
}
template <class t>
void print(t x, ll suc = 1) {
  cout << x;
  if (suc == 1) cout << "\n";
  if (suc == 2) cout << " ";
}
ll read() {
  ll i;
  cin >> i;
  return i;
}
vi readvi(ll n, ll off = 0) {
  vi v(n);
  for (ll i = ll(0); i < ll(n); i++) v[i] = read() + off;
  return v;
}
template <class T>
void print(const vector<T>& v, ll suc = 1) {
  for (ll i = ll(0); i < ll(v.size()); i++)
    print(v[i], i == ll(v.size()) - 1 ? suc : 2);
}
string readString() {
  string s;
  cin >> s;
  return s;
}
template <class T>
T sq(const T& t) {
  return t * t;
}
void yes(bool ex = true) {
  cout << "Yes"
       << "\n";
  if (ex) exit(0);
}
void no(bool ex = true) {
  cout << "No"
       << "\n";
  if (ex) exit(0);
}
void possible(bool ex = true) {
  cout << "Possible"
       << "\n";
  if (ex) exit(0);
}
void impossible(bool ex = true) {
  cout << "Impossible"
       << "\n";
  if (ex) exit(0);
}
constexpr ll ten(ll n) { return n == 0 ? 1 : ten(n - 1) * 10; }
const ll infLL = LLONG_MAX / 3;
const ll inf = infLL;
ll topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
ll topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
ll botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
ll botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
ll popcount(signed t) { return __builtin_popcount(t); }
ll popcount(ll t) { return __builtin_popcountll(t); }
bool ispow2(ll i) { return i && (i & -i) == i; }
ll mask(ll i) { return (ll(1) << i) - 1; }
bool inc(ll first, ll second, ll c) { return first <= second && second <= c; }
template <class t>
void mkuni(vc<t>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
ll rand_int(ll l, ll r) {
  static random_device rd;
  static mt19937_64 gen(rd());
  return uniform_int_distribution<ll>(l, r)(gen);
}
template <class t>
ll lwb(const vc<t>& v, const t& first) {
  return lower_bound(v.begin(), v.end(), first) - v.begin();
}
template <class N, class E>
struct treedp {
  const vvc<E>& g;
  vc<N> first, second, res;
  void dfs1(ll v, ll p) {
    first[v].init(v);
    E up;
    for (auto e : g[v])
      if (e != p) {
        dfs1(e, v);
        second[e] = first[v];
        first[v] = first[v] + first[e];
      } else
        up = e;
    if (p != -1) first[v] = first[v].up(up);
  }
  void dfs2(ll v, ll p, N cur) {
    for (ll i = ll(g[v].size()) - 1; i >= ll(0); i--) {
      auto e = g[v][i];
      if (e == p) continue;
      dfs2(e, v, (cur + second[e]).up(e));
      cur = cur + first[e];
    }
    res[v] = cur;
  }
  treedp(const vvc<E>& gg)
      : g(gg), first(g.size()), second(g.size()), res(g.size()) {
    dfs1(0, -1);
    N tmp;
    tmp.init(0);
    dfs2(0, -1, tmp);
  }
};
struct E {
  ll to, cost;
  operator ll() const { return to; }
};
vi wt;
struct N {
  ll top, c, s;
  void init(ll v) {
    top = c = wt[v];
    s = 0;
  }
  N up(E e) const { return N{wt[e.to], c + wt[e.to], s + c * e.cost}; }
  N operator+(const N& r) const {
    assert(top == r.top);
    return N{top, c + r.c - top, s + r.s};
  }
};
const ll vmax = 5010;
bool pri[vmax];
void initprime() {
  for (ll i = ll(2); i < ll(vmax); i++) pri[i] = 1;
  for (ll i = ll(2); i < ll(vmax); i++)
    if (pri[i])
      for (ll j = i * 2; j < vmax; j += i) pri[j] = 0;
}
vi factors[vmax];
void initfactor() {
  for (ll i = ll(2); i < ll(vmax); i++)
    if (pri[i])
      for (ll j = i; j < vmax; j += i) {
        ll x = j;
        while (x % i == 0) {
          factors[j].push_back(i);
          x /= i;
        }
      }
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  initprime();
  initfactor();
  vi cnt(vmax);
  {
    ll n;
    cin >> n;
    if (n < 0) {
      n = -n;
      for (ll i = ll(0); i < ll(n); i++) cnt[max(i, ll(1))]++;
    } else {
      for (ll _ = ll(0); _ < ll(n); _++) {
        cnt[max(ll(1), read())]++;
      }
    }
  }
  vi ps;
  for (ll i = ll(0); i < ll(vmax); i++)
    if (pri[i]) ps.push_back(i);
  ll s = ps.size();
  auto getidx = [&](ll p) { return s - 1 - lwb(ps, p); };
  void(0);
  auto lca = [&](const vi& first, const vi& second) {
    for (ll i = ll(0); i < ll(s); i++)
      if (first[i] != second[i]) {
        vi res(s);
        for (ll j = ll(0); j < ll(i); j++) res[j] = first[j];
        res[i] = min(first[i], second[i]);
        return res;
      }
    return first;
  };
  auto dist = [&](const vi& first, const vi& second) {
    ll res = 0;
    for (ll i = ll(0); i < ll(s); i++) res += second[i] - first[i];
    return res;
  };
  auto asds = [&](const vi& first, const vi& second) {
    for (ll i = ll(0); i < ll(s); i++)
      if (first[i] != second[i]) {
        if (first[i] > second[i]) return false;
        for (ll j = ll(i + 1); j < ll(s); j++)
          if (first[j]) return false;
        return true;
      }
    void(0);
    assert(false);
  };
  ll n = 0;
  vc<tuple<ll, ll, ll>> es;
  vc<pair<vi, ll>> st;
  auto ae = [&](ll i) {
    es.emplace_back(st[i].second, st[i + 1].second,
                    dist(st[i].first, st[i + 1].first));
  };
  vi cur(s);
  for (ll i = ll(0); i < ll(vmax); i++) {
    for (auto f : factors[i]) cur[getidx(f)]++;
    if (cnt[i]) {
      void(0);
      void(0);
      if (st.size()) {
        auto z = lca(st.back().first, cur);
        void(0);
        auto itr = lower_bound(st.begin(), st.end(), pair<vi, ll>(z, -1));
        if (itr == st.end() || itr->first != z) {
          st.insert(itr, make_pair(z, n++));
          wt.push_back(0);
        }
        while (!asds(st.back().first, cur)) {
          ae(ll(st.size()) - 2);
          st.pop_back();
        }
      }
      st.emplace_back(cur, n++);
      wt.push_back(cnt[i]);
    }
  }
  for (ll i = ll(0); i < ll(ll(st.size()) - 1); i++) ae(i);
  {
    vvc<E> tr(n);
    for (auto e : es) {
      ll first, second, c;
      tie(first, second, c) = e;
      tr[first].push_back({second, c});
      tr[second].push_back({first, c});
    }
    auto dp = treedp<N, E>(tr).res;
    ll ans = inf;
    for (ll i = ll(0); i < ll(n); i++) chmin(ans, dp[i].s);
    cout << ans << endl;
  }
}