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.
//package contest_488;

import java.util.*;
import java.io.*;
public class q1
{
	public class Node implements Comparable<Node>
	{
		int a,b;
		public Node(int a, int b)
		{
			this.a=a;
			this.b=b;
		}
		@Override
		public int compareTo(Node o)
		{
			return Integer.compare(this.a, o.a);
		}
	}
public static void main(String[] args)
{
    InputReader in=new InputReader(System.in);
   OutputWriter out=new OutputWriter(System.out);
   Scanner sc=new Scanner(System.in);
    q1 obj=new q1();
   	int n=in.readInt();
   	int a=(int)Math.sqrt(n);
   	int b=n/a;
   	int x=1;
   	int count=0;
   	while(x<=b)
   	{
   		int i=(n+1)-x*a;
   		for(int k=1;k<=a;k++,i++)
   		{
   			count++;
   			out.print(i+" ");
   		}
   		x++;
   		
   	}
   	for(int i=1;count<n;i++,count++)
   	{
   		out.print(i+" ");
   	}
   	out.flush();
   	out.close();
}
}
class InputReader {

    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    private SpaceCharFilter filter;

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

    public int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    public int readInt() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int res = 0;
        do {
            if (c < '0' || c > '9')
                throw new InputMismatchException();
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }

    public String readString() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    public boolean isSpaceChar(int c) {
        if (filter != null)
            return filter.isSpaceChar(c);
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public String next() {
        return readString();
    }

    public interface SpaceCharFilter {
        public boolean isSpaceChar(int ch);
    }
}

class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
        this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }

    public void printLine(Object... objects) {
        print(objects);
        writer.println();
    }

    public void close() {
        writer.close();
    }

    public void flush() {
        writer.flush();
    }

}

class IOUtils {

    public static int[] readIntArray(InputReader in, int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; i++)
            array[i] = in.readInt();
        return array;
    }

}

