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.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;
public class C502 {

	public static void main(String[] args) {
		FastScanner in = new FastScanner(System.in);
		int N = in.nextInt();
		int[] ar = new int[N];
		int s = (int) Math.sqrt(N);
		int index = N - s;
		int num = 1;
		while(num <= N)
		{
			for(int i = 0; num <= N && i < s; i++)
				ar[index++] = num++;
			index -= 2 * s;
			while(index < 0) index++;
		}
		StringBuilder sb = new StringBuilder();
		for(int i : ar)
			sb.append(" " + i);
		System.out.println(sb.toString().trim());
	}
	
	/**
	 * Source: Matt Fontaine
	 */
	public static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

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

		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++];
		}

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

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

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

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

}
