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.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Collection;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.Deque;
import java.util.ArrayDeque;
import java.io.InputStream;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author MaxHeap
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskC solver = new TaskC();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskC {

    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt();
      int[] per = new int[n];
      for (int i = 0; i < n; i++) {
        per[i] = i + 1;
      }
//    ArrayUtils.reverse(per);
      int block = (int) Math.sqrt(n);
      boolean[] taken = new boolean[n];
      Deque<Integer> q = new ArrayDeque<>();
      int i = block - 1;
      int blockCounter = 1;
      while (i < n) {
        for (int j = i, cp = 0; cp < block && j >= 0; j--, ++cp) {
          q.addLast(per[j]);
          taken[j] = true;
        }
        ++blockCounter;
        i = Math.min(n, blockCounter * block - 1);
      }
      for (int j = n - 1; j >= 0; j--) {
        if (!taken[j]) {
          q.addLast(per[j]);
        }
      }
      while (!q.isEmpty()) {
        out.print(q.pollLast() + " ");
      }
      out.println();
//    int best = (int) 1e9;
//    do {
//      int normal = lis(per, n);
//      int[] cl = per.clone();
//      ArrayUtils.reverse(cl);
//      int reversed = lis(cl, n);
//      if (best >= normal + reversed) {
//        out.println(normal + reversed);
//        out.println(Arrays.toString(per));
//        best = normal + reversed;
//      }
//    } while (ArrayUtils.nextPermutation(per));
    }

  }

  static class InputReader {

    private InputStream stream;
    private byte[] buf = new byte[1 << 13];
    private int curChar;
    private int numChars;

    public InputReader(InputStream stream) {
      this.stream = stream;
    }

    public int read() {
      if (this.numChars == -1) {
        throw new UnknownError();
      } else {
        if (this.curChar >= this.numChars) {
          this.curChar = 0;

          try {
            this.numChars = this.stream.read(this.buf);
          } catch (IOException ex) {
            throw new InputMismatchException();
          }

          if (this.numChars <= 0) {
            return -1;
          }
        }

        return this.buf[this.curChar++];
      }
    }

    public int nextInt() {
      int c;
      for (c = this.read(); isSpaceChar(c); c = this.read()) {
      }

      byte sgn = 1;
      if (c == 45) {
        sgn = -1;
        c = this.read();
      }

      int res = 0;

      while (c >= 48 && c <= 57) {
        res *= 10;
        res += c - 48;
        c = this.read();
        if (isSpaceChar(c)) {
          return res * sgn;
        }
      }

      throw new InputMismatchException();
    }

    public static boolean isSpaceChar(int c) {
      return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
    }

  }
}

