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;
double PI = acos(-1.0);
long long mod = 998244353;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long prod(long long a, long long b) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1ll) res = (res + a) % mod;
    a = (2ll * a) % mod;
    b >>= 1ll;
  }
  return res;
}
long long power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2ll) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2ll;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> ans(n, 0);
  long long x = sqrt(n);
  long long a = n % x;
  long long b = n / x - a;
  long long p = 0, v = n;
  for (long long i = 0; i < a; i++) {
    for (long long j = p + x; j >= p; j--) {
      ans[j] = v--;
    }
    p += (x + 1);
  }
  for (long long i = 0; i < b; i++) {
    for (long long j = p + x - 1; j >= p; j--) {
      ans[j] = v--;
    }
    p += x;
  }
  for (auto e : ans) {
    cout << e << " ";
  }
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
