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.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.InputMismatchException;

/**
 * @author thesparkboy
 *
 */
public class A3 {

	static void reverse(int arr[], int n, int k) {
		for (int i = 0; i < n; i += k) {
			int left = i;

			int right = Math.min(i + k - 1, n - 1);
			int temp;

			while (left < right) {
				temp = arr[left];
				arr[left] = arr[right];
				arr[right] = temp;
				left += 1;
				right -= 1;
			}
		}
	}

	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		OutputWriter out = new OutputWriter(System.out);
		int n = in.readInt();
		long ans = 0;
		int arr[] = new int[n];
		StringBuilder sb = new StringBuilder();
		int slot = (int)Math.sqrt(n);
		int ap = (n + 2) / 2;
		for (int i = 0; i < n; i++) {
			arr[i] = n - i;
		}
		if (slot > 0)
			reverse(arr, n, slot);
		for (int i = 0; i < n; i++) {
			sb.append(arr[i] + " ");
		}
		System.out.println(sb);
		out.close();
	}

	private static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		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 long readLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			long sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long 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 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();
			StringBuffer res = new StringBuffer();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

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

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

	private static class OutputWriter {
		private final PrintWriter writer;

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