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;
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class A, class B, class C>
struct Triple {
  A first;
  B second;
  C third;
  bool operator<(const Triple& t) const {
    if (first != t.first) return first < t.first;
    if (second != t.second) return second < t.second;
    return third < t.third;
  }
};
template <class T>
void ResizeVec(T&, vector<long long>) {}
template <class T>
void ResizeVec(vector<T>& vec, vector<long long> sz) {
  vec.resize(sz[0]);
  sz.erase(sz.begin());
  if (sz.empty()) {
    return;
  }
  for (T& v : vec) {
    ResizeVec(v, sz);
  }
}
template <class A, class B, class C>
ostream& operator<<(ostream& out, Triple<A, B, C> t) {
  return out << "(" << t.first << ", " << t.second << ", " << t.third << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  if (0) cout << fixed << setprecision(10);
  cin.tie(0);
  long long n;
  cin >> n;
  long long I = 1;
  long long D = n;
  for (long long i = (1); i <= (n); ++i) {
    long long d = (n + i - 1) / i;
    if (i + d < I + D) {
      I = i;
      D = d;
    }
  }
  vector<long long> ans(n);
  iota((ans).begin(), (ans).end(), 1);
  for (long long i = 0; i < (I); ++i) {
    long long b = i * D;
    long long e = min(n, (i + 1) * D);
    reverse(ans.begin() + b, ans.begin() + e);
  }
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << endl;
  return 0;
}
