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;
struct Scanner {
  bool hasNext = 1;
  bool hasRead = 1;
  int nextInt() {
    hasRead = 0;
    int res = 0;
    char flag = 1, ch = getchar();
    while (ch != EOF && !isdigit(ch)) {
      hasRead = 1;
      flag = (ch == '-') ? -flag : flag;
      ch = getchar();
    }
    while (ch != EOF && isdigit(ch)) {
      hasRead = 1;
      res = res * 10 + (ch - '0');
      ch = getchar();
    }
    if (ch == EOF) hasNext = 0;
    return res * flag;
  }
  long long nextLL() {
    hasRead = 0;
    long long res = 0;
    char flag = 1, ch = getchar();
    while (ch != EOF && !isdigit(ch)) {
      hasRead = 1;
      flag = (ch == '-') ? -flag : flag;
      ch = getchar();
    }
    while (ch != EOF && isdigit(ch)) {
      hasRead = 1;
      res = res * 10 + (ch - '0');
      ch = getchar();
    }
    if (ch == EOF) hasNext = 0;
    return res * flag;
  }
  char nextChar() {
    hasRead = 0;
    char ch = getchar();
    while (ch != EOF && isspace(ch)) {
      hasRead = 1;
      ch = getchar();
    }
    if (ch == EOF) hasNext = 0;
    return ch;
  }
  int nextString(char *str) {
    hasRead = 0;
    int len = 0;
    char ch = getchar();
    while (ch != EOF && isspace(ch)) {
      hasRead = 1;
      ch = getchar();
    }
    while (ch != EOF && !isspace(ch)) {
      hasRead = 1;
      str[++len] = ch;
      ch = getchar();
    }
    str[len + 1] = 0;
    if (ch == EOF) hasNext = 0;
    return len;
  }
} sc;
long long rd() {
  long long x = sc.nextLL();
  return x;
}
void rd(int &x) { x = sc.nextInt(); }
void rd(long long &x) { x = sc.nextLL(); }
void rd(char &x) { x = sc.nextChar(); }
void rd(char *x) { sc.nextString(x); }
template <typename T1, typename T2>
void rd(pair<T1, T2> &x) {
  rd(x.first);
  rd(x.second);
}
template <typename T>
void rd(T *x, int n) {
  for (int i = 1; i <= n; ++i) rd(x[i]);
}
void printInt(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) printInt(x / 10);
  putchar('0' + x % 10);
}
void printLL(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) printLL(x / 10);
  putchar('0' + x % 10);
}
void pr(int x, char ch = '\n') {
  printInt(x);
  putchar(ch);
}
void pr(long long x, char ch = '\n') {
  printLL(x);
  putchar(ch);
}
template <typename T1, typename T2>
void pr(pair<T1, T2> x, char ch = '\n') {
  pr(x.first, ' ');
  pr(x.second, ch);
}
template <typename T>
void pr(T *x, int n) {
  for (int i = 1; i <= n; ++i) pr(x[i], " \n"[i == n]);
}
template <typename T>
void pr(vector<T> &x) {
  int n = x.size();
  for (int i = 1; i <= n; ++i) pr(x[i - 1], " \n"[i == n]);
}
template <typename T>
void cmin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void cmax(T &x, T y) {
  if (y > x) x = y;
}
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1.0);
const double EPS = 1e-9;
const int MOD = 998244353;
int n;
int a[400005];
int m;
pair<int, int> p[400005];
int cnt[400005];
int suf[400005];
vector<int> b[400005];
int go() {
  while (m >= 1 && p[m].first == 0) --m;
  assert(m >= 1);
  --p[m].first;
  return p[m].second;
}
struct Solver {
  void InitOnce() {}
  void Read() {
    rd(n);
    rd(a, n);
  }
  void Solve() {
    sort(a + 1, a + 1 + n);
    m = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == p[m].second)
        ++p[m].first;
      else {
        p[++m].second = a[i];
        p[m].first = 1;
      }
    }
    sort(p + 1, p + 1 + m);
    for (int i = 1; i <= n + 1; ++i) cnt[i] = 0, suf[i] = 0;
    for (int i = 1; i <= m; ++i) ++cnt[p[i].first];
    for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + cnt[i];
    int w = 0, h = 0, cursum = 0;
    for (int i = 1; i * i <= n; ++i) {
      cursum += cnt[i] * i;
      int tmp = cursum + suf[i + 1] * i;
      int tw = i;
      int th = tmp / tw;
      if (tw > th) continue;
      tmp = tw * th;
      if (tmp > w * h) {
        w = tw;
        h = th;
      }
    }
    pr(w * h);
    printf("%d %d\n", h, w);
    for (int i = 0; i < h; ++i) {
      b[i].clear();
      b[i].resize(w);
    }
    for (int i = m; i >= 1; --i) {
      if (p[i].first > w) p[i].first = w;
    }
    for (int i = 0; i < h; ++i) {
      for (int j = 0; j < w; ++j) b[(i + j) % h][j] = go();
    }
    for (int i = 0; i < h; ++i) {
      for (int j = 0; j < w; ++j) printf("%d ", b[i][j]);
      printf("\n");
    }
  }
} solver;
int main() {
  solver.InitOnce();
  int t = 1;
  t = INF;
  while (t--) {
    solver.Read();
    if (!sc.hasRead) break;
    solver.Solve();
    if (!sc.hasNext) break;
  }
  return 0;
}