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.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;

public class ThePhoneNumber {
	static int mod = 1000000007;
	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {

		int n=in.nextInt();
		int mini=mod;
		int ans=mod;
		for(int i=1;i<=n;i++)
		{
			if(i+ceil((double)n/i)<ans)
			{
				ans=(int) (i+ceil((double)n/i));
				mini=i;
			}
		}
		for(int i=1;i<=n;i++)
		{
			int[] a=new int[mini];
			int c=0;
			while(i<=n && c<mini)
			{
				a[c]=i;
				i++;
				c++;
			}
			i--;
			for(int j=c-1;j>=0;j--)
			{
				out.print(a[j]+" ");
			}
		}
		out.close();

	}

	static class InputReader {
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;

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

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

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				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 String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

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

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}

	private static void tr(Object... o) {
		if (!(System.getProperty("ONLINE_JUDGE") != null))
			System.out.println(Arrays.deepToString(o));
	}
}


