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.util.*;
import java.io.*;

public class File {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		
		if (n == 1) {
			System.out.println(1);
			return;
		}
		else if (n == 2) {
			System.out.println(1 + " " + 2);
			return;
		}
						
		// All increasing or all decreasing.
		int min = n + 1;
		int groupSize = n; // Best size of one increasing group of numbers.
		int groupCount = 1; // Best number of groups.
		
		// Answer is smallest sum equal to:
		// (# groups) + (max size of a group).
		for (int size = n; size >= 1; size--) {
			int groups = (int)Math.ceil((double)n / (double)size);
		
			if (size + groups < min) {
				min = size + groups;
				groupSize = size;
				groupCount = groups;
			}
		}
		
		while (n > 0) {
			int m = Math.min(n, groupSize);
			
			for (int i = n - m + 1; i <= n; i++) {
				System.out.print(i + " ");
			}
			
			n -= m;
		}
	}
}