#include <bits/stdc++.h>
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
using namespace std;
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void printvector(std::vector<long long> v) {
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
long long findlcm(vector<long long> arr, long long n) {
  long long ans = arr[0];
  for (int i = 1; i < n; i++) ans = (((arr[i] * ans)) / (Gcd(arr[i], ans)));
  return ans;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
vector<long long> primeFactors(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
long long C[2001][2000];
void binomialCoeff(long long n, long long k) {
  long long i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= k; j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 1000000007 + C[i - 1][j] % 1000000007) %
                  1000000007;
    }
  }
}
long long spf[100010];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 100010; i++) spf[i] = i;
  for (int i = 4; i < 100010; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 100010; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 100010; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
const long long N = 100000;
long long n;
long long tree[2 * N];
void build(long long arr[]) {
  for (long long i = 0; i < n; i++) {
    tree[n + i] = arr[i];
  }
  for (long long i = n - 1; i > 0; --i) {
    tree[i] = tree[i << 1] + tree[i << 1 | 1];
  }
}
void updateTreeNode(long long p, long long value) {
  tree[p + n] = value;
  p = p + n;
  for (long long i = p; i > 1; i >>= 1) {
    tree[i >> 1] = tree[i] + tree[i ^ 1];
  }
}
long long query(long long l, long long r) {
  long long res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      res += tree[l++];
    }
    if (r & 1) {
      res += tree[--r];
    }
  }
  return res;
}
vector<long long> shortPath(vector<pair<long long, long long> > v[],
                            long long src, long long V = 0) {
  set<pair<long long, long long> > setds;
  long long sizee = V;
  long long IN = 0x3f3f3f3f;
  vector<long long> dist(sizee, IN);
  setds.insert(make_pair(0, src));
  dist[src] = 0;
  while (!setds.empty()) {
    pair<long long, long long> tmp = *(setds.begin());
    setds.erase(setds.begin());
    long long u = tmp.second;
    vector<pair<long long, long long> >::iterator i;
    for (i = v[u].begin(); i != v[u].end(); ++i) {
      long long v = (*i).first;
      long long weight = (*i).second;
      if (dist[v] > dist[u] + weight) {
        if (dist[v] != 0x3f3f3f3f)
          setds.erase(setds.find(make_pair(dist[v], v)));
        dist[v] = dist[u] + weight;
        setds.insert(make_pair(dist[v], v));
      }
    }
  }
  return dist;
}
void dfs(std::vector<long long> v[], long long s, bool visited[]) {
  visited[s] = true;
  for (long long i = 0; i < v[s].size(); ++i) {
    if (!visited[v[s][i]]) {
      dfs(v, v[s][i], visited);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  long long ar[n];
  long long br[n];
  for (int i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
    br[i] = ar[i];
  }
  sort(br, br + n);
  long long ind = n - 1;
  long long min = br[n - 1] - br[0];
  long long j = 0;
  long long jj = 1;
  long long cr[n];
  if (n % 2 == 0) {
    for (int i = 0, _n = (n / 2); i < _n; i++) {
      cr[i] = br[j];
      j++;
      cr[n - 1 - i] = br[j];
      j++;
    }
  } else {
    for (int i = 0, _n = ((n / 2)); i < _n; i++) {
      cr[i] = br[j];
      j++;
      cr[n - 1 - i] = br[j];
      j++;
    }
    cr[n / 2] = br[j];
  }
  for (int i = 0, _n = (n); i < _n; i++) {
    cout << cr[i] << " ";
  }
}

### Prompt

Create a solution in CPP for the following problem:

Cowboy Vlad has a birthday today! There are n children who came to the celebration. In order to greet Vlad, the children decided to form a circle around him. Among the children who came, there are both tall and low, so if they stand in a circle arbitrarily, it may turn out, that there is a tall and low child standing next to each other, and it will be difficult for them to hold hands. Therefore, children want to stand in a circle so that the maximum difference between the growth of two neighboring children would be minimal possible.

Formally, let's number children from 1 to n in a circle order, that is, for every i child with number i will stand next to the child with number i+1, also the child with number 1 stands next to the child with number n. Then we will call the discomfort of the circle the maximum absolute difference of heights of the children, who stand next to each other.

Please help children to find out how they should reorder themselves, so that the resulting discomfort is smallest possible.

Input

The first line contains a single integer n (2 ≤ n ≤ 100) — the number of the children who came to the cowboy Vlad's birthday.

The second line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) denoting heights of every child.

Output

Print exactly n integers — heights of the children in the order in which they should stand in a circle. You can start printing a circle with any child.

If there are multiple possible answers, print any of them.

Examples

Input


5
2 1 1 3 2


Output


1 2 3 2 1


Input


3
30 10 20


Output


10 20 30

Note

In the first example, the discomfort of the circle is equal to 1, since the corresponding absolute differences are 1, 1, 1 and 0. Note, that sequences [2, 3, 2, 1, 1] and [3, 2, 1, 1, 2] form the same circles and differ only by the selection of the starting point.

In the second example, the discomfort of the circle is equal to 20, since the absolute difference of 10 and 30 is equal to 20.

### Response

```cpp
#include <bits/stdc++.h>
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
using namespace std;
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void printvector(std::vector<long long> v) {
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
long long findlcm(vector<long long> arr, long long n) {
  long long ans = arr[0];
  for (int i = 1; i < n; i++) ans = (((arr[i] * ans)) / (Gcd(arr[i], ans)));
  return ans;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
vector<long long> primeFactors(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
long long C[2001][2000];
void binomialCoeff(long long n, long long k) {
  long long i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= k; j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 1000000007 + C[i - 1][j] % 1000000007) %
                  1000000007;
    }
  }
}
long long spf[100010];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 100010; i++) spf[i] = i;
  for (int i = 4; i < 100010; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 100010; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 100010; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
const long long N = 100000;
long long n;
long long tree[2 * N];
void build(long long arr[]) {
  for (long long i = 0; i < n; i++) {
    tree[n + i] = arr[i];
  }
  for (long long i = n - 1; i > 0; --i) {
    tree[i] = tree[i << 1] + tree[i << 1 | 1];
  }
}
void updateTreeNode(long long p, long long value) {
  tree[p + n] = value;
  p = p + n;
  for (long long i = p; i > 1; i >>= 1) {
    tree[i >> 1] = tree[i] + tree[i ^ 1];
  }
}
long long query(long long l, long long r) {
  long long res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      res += tree[l++];
    }
    if (r & 1) {
      res += tree[--r];
    }
  }
  return res;
}
vector<long long> shortPath(vector<pair<long long, long long> > v[],
                            long long src, long long V = 0) {
  set<pair<long long, long long> > setds;
  long long sizee = V;
  long long IN = 0x3f3f3f3f;
  vector<long long> dist(sizee, IN);
  setds.insert(make_pair(0, src));
  dist[src] = 0;
  while (!setds.empty()) {
    pair<long long, long long> tmp = *(setds.begin());
    setds.erase(setds.begin());
    long long u = tmp.second;
    vector<pair<long long, long long> >::iterator i;
    for (i = v[u].begin(); i != v[u].end(); ++i) {
      long long v = (*i).first;
      long long weight = (*i).second;
      if (dist[v] > dist[u] + weight) {
        if (dist[v] != 0x3f3f3f3f)
          setds.erase(setds.find(make_pair(dist[v], v)));
        dist[v] = dist[u] + weight;
        setds.insert(make_pair(dist[v], v));
      }
    }
  }
  return dist;
}
void dfs(std::vector<long long> v[], long long s, bool visited[]) {
  visited[s] = true;
  for (long long i = 0; i < v[s].size(); ++i) {
    if (!visited[v[s][i]]) {
      dfs(v, v[s][i], visited);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  long long ar[n];
  long long br[n];
  for (int i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
    br[i] = ar[i];
  }
  sort(br, br + n);
  long long ind = n - 1;
  long long min = br[n - 1] - br[0];
  long long j = 0;
  long long jj = 1;
  long long cr[n];
  if (n % 2 == 0) {
    for (int i = 0, _n = (n / 2); i < _n; i++) {
      cr[i] = br[j];
      j++;
      cr[n - 1 - i] = br[j];
      j++;
    }
  } else {
    for (int i = 0, _n = ((n / 2)); i < _n; i++) {
      cr[i] = br[j];
      j++;
      cr[n - 1 - i] = br[j];
      j++;
    }
    cr[n / 2] = br[j];
  }
  for (int i = 0, _n = (n); i < _n; i++) {
    cout << cr[i] << " ";
  }
}

```