Problem: You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.

A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.

What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.

Input

The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).

Output

In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.

Examples

Input


12
3 1 4 1 5 9 2 6 5 3 5 8


Output


12
3 4
1 2 3 5
3 1 5 4
5 6 8 9


Input


5
1 1 1 1 1


Output


1
1 1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll llinf = (1ll << 61) - 1;
const double eps = 1e-6, ldeps = 1e-9;
struct Dbg {
  static constexpr auto &os = cout;
  template <class C>
  static auto dud(C *x) -> decltype(os << *x, 0);
  template <class C>
  static char dud(...);
  template <class C>
  typename enable_if<sizeof dud<C>(0) != 1, Dbg &>::type operator<<(
      const C &x) {
    os << x;
    return *this;
  }
  template <class C>
  Dbg &dump(C b, C e) {
    *this << "\n[";
    int cur = 0;
    for (C i = b; i != e and cur < 26; i++, cur++)
      *this << ", " + 2 * (i == b) << *i;
    return *this << ']';
  }
  template <class C>
  typename enable_if<sizeof dud<C>(0) == 1, Dbg &>::type operator<<(
      const C &x) {
    return dump(begin(x), end(x));
  }
  template <class C, size_t X>
  typename enable_if<!is_same<C, const char>::value, Dbg &>::type operator<<(
      C (&x)[X]) {
    return dump(begin(x), end(x));
  }
  template <class B, class C>
  Dbg &operator<<(const pair<B, C> &x) {
    return *this << '{' << x.first << ", " << x.second << '}';
  }
  template <class C, size_t Y>
  struct TP {
    void operator()(Dbg &os, const C &t) {
      TP<C, Y - 1>()(os, t);
      os << ", " << get<Y - 1>(t);
    }
  };
  template <class C>
  struct TP<C, 1> {
    void operator()(Dbg &os, const C &t) { os << get<0>(t); }
  };
  template <class... C>
  Dbg &operator<<(const tuple<C...> &t) {
    os << '{', TP<decltype(t), sizeof...(C)>()(*this, t);
    return *this << '}';
  }
  Dbg &operator<<(ostream &(*x)(std::ostream &)) {
    os << x;
    return *this;
  }
} dbg;
void err(istringstream *iss) { delete iss; }
template <class C, class... Args>
void err(istringstream *iss, const C &val, const Args &...args) {
  string name;
  *iss >> name;
  if (name.back() == ',') name.pop_back();
  dbg << name << " = " << val << "; ", err(iss, args...);
}
int T, Q, cn;
const int inf = 1000000007, mxn = 400005;
int ori[mxn];
vector<int> a;
template <class C>
C compress(vector<C> &v) {
  int sz = int(v.size());
  vector<pair<C, int> > tmp(sz);
  for (int i = 0; i < sz; i++) {
    tmp[i] = {v[i], i};
  }
  sort(begin(tmp), end(tmp));
  int cur = 1;
  for (int i = 0; i < sz; i++) {
    cur += i and tmp[i].first != tmp[i - 1].first;
    ori[cur] = tmp[i].first;
    v[tmp[i].second] = cur;
  }
  return cur;
}
int n, cnt[mxn], cntcnt[mxn], cs[mxn];
struct cmp {
  bool operator()(const int &ls, const int &rs) const {
    return tie(cnt[ls], ls) > tie(cnt[rs], rs);
  }
};
void build(int r, int c) {
  vector<int> tmp;
  for (int i = 0; i <= n; i++) {
    int o = min(r, cnt[i]);
    for (int j = 0; j < o; j++) {
      tmp.emplace_back(i);
    }
  }
  sort(begin(tmp), end(tmp), cmp());
  while (int(tmp.size()) > r * c) {
    tmp.pop_back();
  }
  vector<vector<int> > ans(r, vector<int>(c));
  int cur = 0;
  for (int i = 0; i < c; i++) {
    for (int j = 0; j < r; j++) {
      ans[j][(i + j) % c] = ori[tmp[cur++]];
    }
  }
  for (int i = 0; i < r; i++) {
    for (int j = 0; j < c; j++) {
      cout << ans[i][j] << " \n"[j == c - 1];
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.precision(11);
  cin >> n;
  a.resize(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i - 1];
  }
  compress(a);
  for (int i = 0; i < n; i++) {
    cnt[a[i]]++;
  }
  for (int i = 0; i <= n; i++) {
    cntcnt[cnt[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    cs[i] = cs[i - 1] + cntcnt[i];
  }
  ll o = 0;
  int mx = 0, mxi = 0, mxc = 0;
  for (int i = 1; i <= n; i++) {
    o += cs[n] - cs[i - 1];
    if (o >= i * ll(i) and i * (o / i) > mx)
      mxc = int(o / i), mx = max(mx, i * mxc), mxi = i;
  }
  cout << mx << '\n' << mxi << ' ' << mxc << '\n';
  build(mxi, mxc);
}