Problem: Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.

The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.

The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.

Output

In the first line, print a single integer k — the size of the subset.

In the second line, print k integers — the elements of the subset in any order.

If there are multiple optimal subsets, print any.

Examples

Input

4
1 2 3 12


Output

3
1 2 12 


Input

4
1 1 2 2


Output

3
1 1 2 


Input

2
1 2


Output

2
1 2

Note

In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.

In the second case, the optimal subset is <image>. Note that repetition is allowed.

In the last case, any subset has the same median and mean, so all have simple skewness of 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const double EPS = 1e-9, PI = acos(-1);
const double pi = 3.141592653589793238462643383279;
long long GCD(long long a, long long b) { return (b == 0) ? a : GCD(b, a % b); }
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }
template <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
using modint = ModInt<1000000007>;
template <typename T>
struct Combination {
  vector<T> _fact, _rfact, _inv;
  Combination(int sz) : _fact(sz + 1), _rfact(sz + 1), _inv(sz + 1) {
    _fact[0] = _rfact[sz] = _inv[0] = 1;
    for (int i = 1; i <= sz; i++) _fact[i] = _fact[i - 1] * i;
    _rfact[sz] /= _fact[sz];
    for (int i = sz - 1; i >= 0; i--) _rfact[i] = _rfact[i + 1] * (i + 1);
    for (int i = 1; i <= sz; i++) _inv[i] = _rfact[i] * _fact[i - 1];
  }
  inline T fact(int k) const { return _fact[k]; }
  inline T rfact(int k) const { return _rfact[k]; }
  inline T inv(int k) const { return _inv[k]; }
  T P(int n, int r) const {
    if (r < 0 || n < r) return 0;
    return fact(n) * rfact(n - r);
  }
  T C(int p, int q) const {
    if (q < 0 || p < q) return 0;
    return fact(p) * rfact(q) * rfact(p - q);
  }
  T H(int n, int r) const {
    if (n < 0 || r < 0) return (0);
    return r == 0 ? 1 : C(n + r - 1, r);
  }
};
int n;
vector<long long> x;
long long cums[201010];
long double calc(int i, int len) {
  if (n - len <= i) return -1e9;
  if (i - len < 0) return -1e9;
  long double sum = cums[i + 1] - cums[i - len] + cums[n] - cums[n - len] -
                    x[i] * (2 * len + 1);
  return sum / (long double)(2 * len + 1);
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  cin >> n;
  x.resize(n);
  for (int i = 0; i < (int)n; ++i) cin >> x[i];
  if (n == 1) {
    cout << 1 << endl;
    cout << x[0] << endl;
    return 0;
  } else if (n == 2) {
    cout << 2 << endl;
    cout << x[0] << " " << x[1] << endl;
    return 0;
  }
  sort(x.begin(), x.end());
  for (int i = 0; i < (int)n; ++i) cums[i + 1] = cums[i] + x[i];
  long double mx = -1;
  int anslen = 0;
  int idx = 0;
  for (int i = 1; i < n - 1; i++) {
    long long lb = 0, ub = n;
    while (ub - lb > 1) {
      int mid = (ub + lb) / 2;
      int len = mid;
      if (i - len < 0)
        ub = mid;
      else if (n - len <= i)
        ub = mid;
      else {
        if (calc(i, mid) > calc(i, mid + 1))
          ub = mid;
        else
          lb = mid;
      }
    }
    if (mx < calc(i, ub)) {
      mx = calc(i, ub);
      idx = i;
      anslen = ub;
    }
    if (mx < calc(i, lb)) {
      mx = calc(i, lb);
      idx = i;
      anslen = lb;
    }
  }
  vector<int> ans;
  for (int i = idx - anslen; i <= idx; i++) ans.push_back(x[i]);
  for (int i = n - anslen; i < n; i++) ans.push_back(x[i]);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}