Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of n can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS). 

A subsequence a_{i_1}, a_{i_2}, …, a_{i_k} where 1≤ i_1 < i_2 < … < i_k≤ n is called increasing if a_{i_1} < a_{i_2} < a_{i_3} < … < a_{i_k}. If a_{i_1} > a_{i_2} > a_{i_3} > … > a_{i_k}, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.

For example, if there is a permutation [6, 4, 1, 7, 2, 3, 5], LIS of this permutation will be [1, 2, 3, 5], so the length of LIS is equal to 4. LDS can be [6, 4, 1], [6, 4, 2], or [6, 4, 3], so the length of LDS is 3.

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.

Input

The only line contains one integer n (1 ≤ n ≤ 10^5) — the length of permutation that you need to build.

Output

Print a permutation that gives a minimum sum of lengths of LIS and LDS. 

If there are multiple answers, print any.

Examples

Input

4


Output

3 4 1 2


Input

2


Output

2 1

Note

In the first sample, you can build a permutation [3, 4, 1, 2]. LIS is [3, 4] (or [1, 2]), so the length of LIS is equal to 2. LDS can be ony of [3, 1], [4, 2], [3, 2], or [4, 1]. The length of LDS is also equal to 2. The sum is equal to 4. Note that [3, 4, 1, 2] is not the only permutation that is valid.

In the second sample, you can build a permutation [2, 1]. LIS is [1] (or [2]), so the length of LIS is equal to 1. LDS is [2, 1], so the length of LDS is equal to 2. The sum is equal to 3. Note that permutation [1, 2] is also valid.
#include <bits/stdc++.h>
using namespace std;
char string_in_buffer[(int)260];
void fast_scan(int &first) { scanf("%d", &first); }
void fast_scan(long long &first) { scanf("%lld", &first); }
void fast_scan(unsigned long long &first) { scanf("%llu", &first); }
void fast_scan(double &first) { scanf("%lf", &first); }
void fast_scan(long double &first) { scanf("%Lf", &first); }
void fast_scan(char &first) {
  scanf("%c", &first);
  if (first == '\n') {
    fast_scan(first);
  }
}
void fast_scan(string &first) {
  scanf("%s", string_in_buffer);
  first = string(string_in_buffer);
}
template <class TFirst, class TSecond>
void fast_scan(pair<TFirst, TSecond> &p) {
  fast_scan(p.first);
  fast_scan(p.second);
}
template <class T>
void fast_scan(vector<T> &v) {
  for (auto &first : v) fast_scan(first);
}
void fast_print(const int &first) { printf("%d", first); }
void fast_print(const long long &first) { printf("%lld", first); }
void fast_print(const unsigned long long &first) { printf("%llu", first); }
void fast_print(const double &first) { printf("%.15lf", first); }
void fast_print(const long double &first) { printf("%.15Lf", first); }
void fast_print(const char &first) { printf("%c", first); };
void fast_print(const string &first) { printf("%s", first.c_str()); }
void fast_print(const char v[]) { fast_print((string)v); }
template <class TFirst, class TSecond>
void fast_print(const pair<TFirst, TSecond> &p) {
  fast_print(p.first);
  fast_print(' ');
  fast_print(p.second);
}
template <class T>
void fast_print(const vector<T> &v) {
  if (v.empty()) return;
  fast_print(v[0]);
  for (int i = 1; i < v.size(); i++) {
    fast_print(' ');
    fast_print(v[i]);
  }
}
template <class T>
void fast_print(const vector<vector<T>> &v) {
  if (v.empty()) return;
  fast_print(v[0]);
  for (int i = 1; i < v.size(); i++) {
    fast_print('\n');
    fast_print(v[i]);
  }
}
template <class T>
void fast_print(const T &v) {
  for (const auto &first : v) {
    fast_print(first);
    fast_print(' ');
  }
}
using namespace std;
namespace smart_io {
string print_start = "";
string sep = " ";
bool first_print = false;
void precall_print() {
  fast_print(print_start);
  print_start = "\n";
  first_print = true;
}
void _print(deque<string>) {}
template <class T, class... Args>
void _print(deque<string> names, T kek, Args... args) {
  if (!first_print) {
    fast_print("\n");
  } else {
    first_print = false;
  }
  fast_print(names.front());
  fast_print(" = ");
  fast_print(kek);
  names.pop_front();
  _print(names, args...);
}
}  // namespace smart_io
template <class T>
ostream &operator,(ostream &os, const T &object) {
  if (!smart_io::first_print) {
    fast_print(smart_io::sep);
  } else {
    smart_io::first_print = false;
  }
  fast_print(object);
  return os;
}
template <class T>
istream &operator,(istream &is, T &object) {
  fast_scan(object);
  return is;
}
namespace typedefs {}
namespace numbers_operation {
template <class T>
T floor_mod(T a, T b) {
  if (a % b == 0) return 0;
  if (a >= 0 && b >= 0) return a % b;
  if (a <= 0 && b <= 0) return a % b;
  return abs(b) - (abs(a) % abs(b));
}
}  // namespace numbers_operation
using namespace numbers_operation;
using namespace typedefs;
signed main(signed argc, char *argv[]) {
  int n;
  cin, n;
  int q = 0;
  while ((q + 1) * (q + 1) <= n) {
    q++;
  }
  vector<int> rez(n);
  for (int i = 0; i < (n); i++) {
    rez[i] = i;
  }
  reverse(rez.begin(), rez.end());
  for (int i = 0; i < n; i += q) {
    reverse(rez.begin() + i, rez.begin() + min(n, i + q));
  }
  for (auto &first : rez) {
    first++;
  };
  smart_io::precall_print();
  cout, rez;
}
